From d64bf9b076f47fecaab9840659481ae69d9e95f4 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Thu, 27 Jul 2023 11:30:27 -0600 Subject: [PATCH] Enderton. Recursion theorem and exercise 4.8. --- Bookshelf/Enderton/Set.tex | 223 +++++++++++++++++++++++++++++++++++-- 1 file changed, 216 insertions(+), 7 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index ad31341..cfd8d52 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -6213,7 +6213,7 @@ Show that $<_L$ is a linear ordering on $A \times B$. \section{Recursion on \texorpdfstring{$\omega$}{the Natural Numbers}}% \hyperlabel{sec:recursion-natural-numbers} -\subsection{\sorry{ +\subsection{\unverified{% Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}% \hyperlabel{sub:recursion-theorem-natural-numbers} @@ -6225,9 +6225,173 @@ Show that $<_L$ is a linear ordering on $A \times B$. \end{theorem} +\begin{note} + This proof was written a few days after reading Enderton's proof as a means of + ensuring I remember the main arguments. +\end{note} + \begin{proof} - TODO + Define set + \begin{align*} + H = \{ v \mid & v \text{ is a function with } \\ + & \text{(a) } \dom{v} \subseteq \omega, \\ + & \text{(b) } \ran{v} \subseteq A, \\ + & \text{(c) if } 0 \in \dom{v}, \text{ then } v(0) = a, \text{ and} \\ + & \text{(d) if } n^+ \in \dom{v}, + \text{ then } n \in \dom{v} \text{ and } v(n^+) = F(v(n)) + \}. + \end{align*} + Define a function satisfying properties (a)-(d) above as \textit{acceptable}. + That is, $H$ is the set of all acceptable functions. + Define $h = \bigcup H$. + We prove that (i) $h$ is a \nameref{ref:function}, (ii) $h \in H$, (iii) + $\dom{h} = \omega$, and (iv) $h$ is unique. + + \paragraph{(i)}% + \label{par:recursion-theorem-natural-numbers-i} + + We prove that $h$ is a function. + Consider set + $$S = \{x \in \omega \mid h(x) = y \text{ for at most one value } y\}.$$ + We show (1) that $0 \in S$ and (2) if $n \in S$ then $n^+ \in S$. + + \subparagraph{(1)}% + \label{spar:recursion-theorem-natural-numbers-i-1} + + Suppose $0 \in \dom{h}$. + By construction, there must exist some $y_1 \in A$ and acceptable function + $v_1$ such that $v_1(0) = y_1$. + Suppose there also exists a $y_2 \in A$ and acceptable function $v_2$ such + that $v_2(0) = y_2$. + By property (c), $v_1(0) = a$ and $v_2(0) = a$. + Thus $y_1 = a = y_2$ and $h(0) = a$. + Therefore $0 \in S$. + + \subparagraph{(2)}% + \label{spar:recursion-theorem-natural-numbers-i-2} + + Suppose $n$ and $n^+$ are members of $\dom{h}$. + By construction, there must exist some $y_1 \in A$ and acceptable function + $v_1$ such that $v_1(n^+) = y_1$. + Suppose there also exists a $y_2 \in A$ and acceptable function $v_2$ such + that $v_2(n^+) = y_2$. + By property (d), it follows $n \in \dom{v_1}$, $n \in \dom{v_2}$, + $v_1(n^+) = F(v_1(n))$, and $v_2(n^+) = F(v_2(n))$. + But $n \in S$ meaning there is at most one value $y$ such that + $v_1(n) = y = v_2(n)$. + Thus $F(v_1(n)) = F(y) = F(v_2(n))$ and $h(n^+) = F(y)$. + Therefore $n^+ \in S$. + + \subparagraph{Conclusion}% + + By \nameref{spar:recursion-theorem-natural-numbers-i-1} and + \nameref{spar:recursion-theorem-natural-numbers-i-2}, $S$ is an + \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Since $S = \omega$, it follows $h$ has at most one value for every + $x \in \omega$. + In other words, $h$ is a function. + + \paragraph{(ii)}% + \label{par:recursion-theorem-natural-numbers-ii} + + We now prove $h \in H$, i.e. $h$ is an acceptable function. + It trivially holds that $\dom{h} \subseteq \omega$ and + $\ran{h} \subseteq A$. + Thus we are left with proving properties (c) and (d). + + \subparagraph{(c)}% + + Note $\{\pair{0, a}\}$ is an acceptable function. + Thus $\pair{0, a} \in h$. + By \nameref{par:recursion-theorem-natural-numbers-i}, $h$ is a function. + Therefore $a$ is the only value $h(0)$ takes on. + + \subparagraph{(d)}% + + Suppose $n^+ \in \dom{h}$. + Then there exists some acceptable function $v$ such that + $v(n^+) = h(n^+)$. + By definition of acceptable, $\pair{n, v(n)} \in v$. + Since \nameref{par:recursion-theorem-natural-numbers-i} indicates $h$ is a + function, $n \in \dom{h}$ and $h(n) = v(n)$. + Also by definition of acceptable, $v(n^+) = F(v(n))$. + Therefore $$h(n^+) = v(n^+) = F(v(n)) = F(h(n)).$$ + Hence $h \in H$. + + \paragraph{(iii)}% + \label{par:recursion-theorem-natural-numbers-iii} + + We now prove that $\dom{h} = \omega$. + We show that (1) $0 \in \dom{h}$ and (2) if $n \in \dom{h}$ then + $n^+ \in \dom{h}$. + + \subparagraph{(1)}% + \label{spar:recursion-theorem-natural-numbers-iii-1} + + We note that $\{\pair{0, a}\}$ is an acceptable function. + By construction of $h$, $0 \in \dom{h}$. + + \subparagraph{(2)}% + \label{spar:recursion-theorem-natural-numbers-iii-2} + + Suppose $n \in \dom{h}$. + Since $n \in \dom{h}$ there exists an acceptable function $v$ with + $n \in \dom{v}$. + Define $$v' = v \cup \{\pair{n^+, F(v(n))}\}.$$ + We prove that $v'$ is acceptable: + + \begin{enumerate}[(a)] + \item It trivially holds that $\dom{v'} \subseteq \omega$. + \item It trivially holds that $\ran{v'} \subseteq A$. + \item If $0 \in \dom{v'}$, then $v'(0) = v(0) = a$. + \item + Suppose $m^+ \in \dom{v'}$ for some $m \in \omega$. + If $m^+ = n^+$, then $m = n$ and $v'(m^+) = F(v(m))$ by construction. + If $m^+ \neq n^+$, then $m^+ \in \dom{v}$. + Since $v$ is an acceptable function, + $v'(m^+) = v(m^+) = F(v(m)) = F(v(m^+))$. + \end{enumerate} + Since $v'$ is acceptable, $n^+ \in \dom{h}$. + + \subparagraph{Conclusion}% + + By \nameref{spar:recursion-theorem-natural-numbers-iii-1} and + \nameref{spar:recursion-theorem-natural-numbers-iii-2}, + $\dom{h}$ is an inductive set. + \nameref{sub:theorem-4b} implies $\dom{h} = \omega$. + + \paragraph{(iv)}% + \label{par:recursion-theorem-natural-numbers-iv} + + We now prove $h$ is a unique function. + Let $h_1$ and $h_2$ both satisfy the conclusion of the theorem. + Define $$S = \{n \in \omega \mid h_1(n) = h_2(n)\}.$$ + It suffices to prove $S$ is an inductive set, for \nameref{sub:theorem-4b} + would then imply $S = \omega$, i.e. $h_1$ and $h_2$ agree on all of + $\omega$. + + By definition of an acceptable function, $h_1(0) = a = h_2(0)$ meaning + $0 \in S$. + Next, suppose $n \in S$. + By \nameref{par:recursion-theorem-natural-numbers-iii}, it follows $n^+$ + in $\dom{h_1}$ and $\dom{h_2}$. + Since both $h_1$ and $h_2$ are acceptable, $h_1(n^+) = F(h_1(n))$ and + $h_2(n^+) = F(h_2(n))$. + Since $n \in S$, $h_1(n) = h_2(n)$. + Therefore $h_1$ and $h_2$ coincide with input $n^+$. + Thus $n^+ \in S$. + Hence $S$ is an inductive set. + + \paragraph{Conclusion}% + + By \nameref{par:recursion-theorem-natural-numbers-i}, + \nameref{par:recursion-theorem-natural-numbers-iii}, and + \nameref{par:recursion-theorem-natural-numbers-iv}, it follows $h$ is a + unique function mapping $\omega$ into $A$. + \nameref{par:recursion-theorem-natural-numbers-ii} shows $h$ satisfies the + desired conditions. \end{proof} @@ -6399,18 +6563,19 @@ Prove the converse to \nameref{sub:theorem-4e}: If \end{proof} -\subsection{\sorry{Exercise 4.7}}% +\subsection{\unverified{Exercise 4.7}}% \hyperlabel{sub:exercise-4.7} -Complete part 4 of the proof of the recursion theorem on $\omega$. +Complete part 4 of the proof of the + \nameref{sub:recursion-theorem-natural-numbers}. \begin{proof} - TODO + Refer to \nameref{par:recursion-theorem-natural-numbers-iv}. \end{proof} -\subsection{\sorry{Exercise 4.8}}% +\subsection{\unverified{Exercise 4.8}}% \hyperlabel{sub:exercise-4.8} Let $f$ be a one-to-one function from $A$ into $A$, and assume that @@ -6420,10 +6585,54 @@ Define $h \colon \omega \rightarrow A$ by recursion: h(0) & = c, \\ h(n^+) & = f(h(n)). \end{align*} +Show that $h$ is one-to-one. \begin{proof} - TODO + Let + $$S = \{x \in \omega \mid \forall y, + \left[ h(x) = h(y) \Rightarrow x = y \right]\}.$$ + We prove that (i) $S$ is an \nameref{ref:inductive-set} and (ii) that $h$ is + one-to-one. + + \paragraph{(i)}% + \label{par:exercise-4.8-i} + + We first show that $0 \in S$. + Suppose there exists some $n \in \omega$ such that $h(0) = c = h(n)$. + For the sake of contradiction, suppose $n \neq 0$. + By \nameref{sub:theorem-4c}, there exists some $m$ such that $m^+ = n$. + Then $$h(n) = h(m^+) = f(h(m)) = c.$$ + But $c \in A - \ran{f}$, meaning the previous identity is an impossibility. + Thus $n = 0$, i.e. $0 \in S$. + + Next, suppose $y, n \in \omega$ such that $n \in S$ and $h(n^+) = h(y)$. + We must show $n^+ \in S$. + There are two cases to consider: + + \subparagraph{Case 1}% + + Suppose $y = 0$. + Then $h(y) = h(0) = c = h(n^+) = f(h(n))$. + Since $c \in A - \ran{f}$, $f(h(n)) \neq c$. + Thus $y \neq 0$, a contradiction. + + \subparagraph{Case 2}% + + Suppose $y \neq 0$. + \nameref{sub:theorem-4c} implies there exists some $z \in \omega$ such + that $z^+ = y$. + Thus $$h(n^+) = f(h(n)) = f(h(z)) = h(z^+).$$ + But $f$ is one-to-one meaning $h(n) = h(z)$. + Since $n \in S$, $h(n) = h(z)$ implies $n = z$ which in turn implies + $n^+ = z^+ = y$. + Thus $n^+$ is in $S$. + + \paragraph{(ii)}% + + By \nameref{par:exercise-4.8-i}, $S \subseteq \omega$ is an inductive set. + Then \nameref{sub:theorem-4b} states $S = \omega$. + Hence $h$ is one-to-one. \end{proof}