Enderton. Recursion theorem and exercise 4.8.

finite-set-exercises
Joshua Potter 2023-07-27 11:30:27 -06:00
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@ -6213,7 +6213,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\section{Recursion on \texorpdfstring{$\omega$}{the Natural Numbers}}% \section{Recursion on \texorpdfstring{$\omega$}{the Natural Numbers}}%
\hyperlabel{sec:recursion-natural-numbers} \hyperlabel{sec:recursion-natural-numbers}
\subsection{\sorry{ \subsection{\unverified{%
Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}% Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}%
\hyperlabel{sub:recursion-theorem-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} \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} \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} \end{proof}
@ -6399,18 +6563,19 @@ Prove the converse to \nameref{sub:theorem-4e}: If
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.7}}% \subsection{\unverified{Exercise 4.7}}%
\hyperlabel{sub: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} \begin{proof}
TODO Refer to \nameref{par:recursion-theorem-natural-numbers-iv}.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.8}}% \subsection{\unverified{Exercise 4.8}}%
\hyperlabel{sub:exercise-4.8} \hyperlabel{sub:exercise-4.8}
Let $f$ be a one-to-one function from $A$ into $A$, and assume that 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(0) & = c, \\
h(n^+) & = f(h(n)). h(n^+) & = f(h(n)).
\end{align*} \end{align*}
Show that $h$ is one-to-one.
\begin{proof} \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} \end{proof}