Enderton. Draft recursion theorems and exercises.
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@ -6149,7 +6149,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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natural number is a transitive set.
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\paragraph{(i)}%
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\label{par:theorem-4f-i}
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\hyperlabel{par:theorem-4f-i}
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First, $\emptyset \in T$ since it vacuously holds that a member of a
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member of $\emptyset$ is itself a member of $\emptyset$.
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@ -6171,7 +6171,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\end{proof}
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\subsection{\unverified{Theorem 4G}}%
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\label{sub:theorem-4g}
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\hyperlabel{sub:theorem-4g}
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\begin{theorem}[4G]
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@ -6186,7 +6186,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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member of a natural number is itself a natural number.
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\paragraph{(i)}%
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\label{par:theorem-4g-i}
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\hyperlabel{par:theorem-4g-i}
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First, it vacuously holds that $\emptyset \in T$.
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Next, let $n \in T$.
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@ -6210,6 +6210,46 @@ Show that $<_L$ is a linear ordering on $A \times B$.
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\end{proof}
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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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Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}%
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\hyperlabel{sub:recursion-theorem-natural-numbers}
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\begin{theorem}
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Let $A$ be a set, $a \in A$, and $F \colon A \rightarrow A$.
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Then there exists a unique function $h \colon \omega \rightarrow A$ such that
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$$h(0) = a,$$ and for every $n \in \omega$, $$h(n^+) = F(h(n)).$$
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Theorem 4H}}%
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\hyperlabel{sub:theorem-4h}
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\begin{theorem}[4H]
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Let $\langle N, S, e \rangle$ be a Peano system.
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Then $\langle \omega, \sigma, 0 \rangle$ is isomorphic to
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$\langle N, S, e \rangle$, i.e., there is a function $h$ mapping $\omega$
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one-to-one onto $N$ in a way that preserves the successor operation
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$$h(\sigma(n)) = S(h(n))$$ and the zero element $$h(0) = e.$$
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\section{Exercises 4}%
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\hyperlabel{sec:exercises-4}
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@ -6359,4 +6399,99 @@ 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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\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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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{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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$c \in A - \ran{f}$.
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Define $h \colon \omega \rightarrow A$ by recursion:
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\begin{align*}
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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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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 4.9}}%
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\hyperlabel{sub:exercise-4.9}
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Let $f$ be a function from $B$ into $B$, and assume that $A \subseteq B$.
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We have two possible methods for constructing the "closure" $C$ of $A$ under
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$f$.
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First define $C^*$ to be the intersection of the closed supersets of $A$:
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$$C^* = \bigcap\{X \mid
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A \subseteq X \subseteq B \land \img{f}{X} \subseteq X\}.$$
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Alternatively, we could apply the recursion theorem to obtain the function $h$
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for which
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\begin{align*}
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h(0) & = A, \\
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h(n^+) & = h(n) \cup \img{f}{h(n)}.
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\end{align*}
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Clearly $h(0) \subseteq h(1) \subseteq \cdots$; define $C_*$ to be
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$\bigcup\ran{h}$; in other words $$C_* = \bigcup_{i \in \omega} h(i).$$
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Show that $C^+ = C_*$.
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[\textit{Suggestion}:
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To show that $C^* \subseteq C_*$, show that $\img{f}{C_*} \subseteq C_*$.
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To show that $C_* \subseteq C^*$, use induction to show that
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$h(n) \subseteq C^*$.]
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 4.10}}%
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\hyperlabel{sub:exercise-4.10}
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In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x^2$, and $A$
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is the closed interval $\icc{\frac{1}{2}}{1}$.
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What is the set called $C^*$ and $C_*$?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 4.11}}%
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\hyperlabel{sub:exercise-4.11}
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In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x - 1$, and
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$A = \{0\}$.
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What is the set called $C^*$ and $C_*$?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 4.12}}%
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\hyperlabel{sub:exercise-4.12}
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Formulate an analogue to Exercise 9 for a function
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$f \colon B \times B \rightarrow B$.
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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