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