Enderton. Draft recursion theorems and exercises.

finite-set-exercises
Joshua Potter 2023-07-25 12:46:26 -06:00
parent 7fe780e72f
commit 526a11a3f6
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@ -6149,7 +6149,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
natural number is a transitive set. natural number is a transitive set.
\paragraph{(i)}% \paragraph{(i)}%
\label{par:theorem-4f-i} \hyperlabel{par:theorem-4f-i}
First, $\emptyset \in T$ since it vacuously holds that a member of a First, $\emptyset \in T$ since it vacuously holds that a member of a
member of $\emptyset$ is itself a member of $\emptyset$. member of $\emptyset$ is itself a member of $\emptyset$.
@ -6171,7 +6171,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof} \end{proof}
\subsection{\unverified{Theorem 4G}}% \subsection{\unverified{Theorem 4G}}%
\label{sub:theorem-4g} \hyperlabel{sub:theorem-4g}
\begin{theorem}[4G] \begin{theorem}[4G]
@ -6186,7 +6186,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
member of a natural number is itself a natural number. member of a natural number is itself a natural number.
\paragraph{(i)}% \paragraph{(i)}%
\label{par:theorem-4g-i} \hyperlabel{par:theorem-4g-i}
First, it vacuously holds that $\emptyset \in T$. First, it vacuously holds that $\emptyset \in T$.
Next, let $n \in T$. Next, let $n \in T$.
@ -6210,6 +6210,46 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof} \end{proof}
\section{Recursion on \texorpdfstring{$\omega$}{the Natural Numbers}}%
\hyperlabel{sec:recursion-natural-numbers}
\subsection{\sorry{
Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}%
\hyperlabel{sub:recursion-theorem-natural-numbers}
\begin{theorem}
Let $A$ be a set, $a \in A$, and $F \colon A \rightarrow A$.
Then there exists a unique function $h \colon \omega \rightarrow A$ such that
$$h(0) = a,$$ and for every $n \in \omega$, $$h(n^+) = F(h(n)).$$
\end{theorem}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Theorem 4H}}%
\hyperlabel{sub:theorem-4h}
\begin{theorem}[4H]
Let $\langle N, S, e \rangle$ be a Peano system.
Then $\langle \omega, \sigma, 0 \rangle$ is isomorphic to
$\langle N, S, e \rangle$, i.e., there is a function $h$ mapping $\omega$
one-to-one onto $N$ in a way that preserves the successor operation
$$h(\sigma(n)) = S(h(n))$$ and the zero element $$h(0) = e.$$
\end{theorem}
\begin{proof}
TODO
\end{proof}
\section{Exercises 4}% \section{Exercises 4}%
\hyperlabel{sec:exercises-4} \hyperlabel{sec:exercises-4}
@ -6359,4 +6399,99 @@ Prove the converse to \nameref{sub:theorem-4e}: If
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.7}}%
\hyperlabel{sub:exercise-4.7}
Complete part 4 of the proof of the recursion theorem on $\omega$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.8}}%
\hyperlabel{sub:exercise-4.8}
Let $f$ be a one-to-one function from $A$ into $A$, and assume that
$c \in A - \ran{f}$.
Define $h \colon \omega \rightarrow A$ by recursion:
\begin{align*}
h(0) & = c, \\
h(n^+) & = f(h(n)).
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.9}}%
\hyperlabel{sub:exercise-4.9}
Let $f$ be a function from $B$ into $B$, and assume that $A \subseteq B$.
We have two possible methods for constructing the "closure" $C$ of $A$ under
$f$.
First define $C^*$ to be the intersection of the closed supersets of $A$:
$$C^* = \bigcap\{X \mid
A \subseteq X \subseteq B \land \img{f}{X} \subseteq X\}.$$
Alternatively, we could apply the recursion theorem to obtain the function $h$
for which
\begin{align*}
h(0) & = A, \\
h(n^+) & = h(n) \cup \img{f}{h(n)}.
\end{align*}
Clearly $h(0) \subseteq h(1) \subseteq \cdots$; define $C_*$ to be
$\bigcup\ran{h}$; in other words $$C_* = \bigcup_{i \in \omega} h(i).$$
Show that $C^+ = C_*$.
[\textit{Suggestion}:
To show that $C^* \subseteq C_*$, show that $\img{f}{C_*} \subseteq C_*$.
To show that $C_* \subseteq C^*$, use induction to show that
$h(n) \subseteq C^*$.]
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.10}}%
\hyperlabel{sub:exercise-4.10}
In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x^2$, and $A$
is the closed interval $\icc{\frac{1}{2}}{1}$.
What is the set called $C^*$ and $C_*$?
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.11}}%
\hyperlabel{sub:exercise-4.11}
In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x - 1$, and
$A = \{0\}$.
What is the set called $C^*$ and $C_*$?
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.12}}%
\hyperlabel{sub:exercise-4.12}
Formulate an analogue to Exercise 9 for a function
$f \colon B \times B \rightarrow B$.
\begin{proof}
TODO
\end{proof}
\end{document} \end{document}