Enderton. Recursion on omega exercises.

finite-set-exercises
Joshua Potter 2023-08-01 12:41:51 -06:00
parent d64bf9b076
commit 8b1da10ef1
1 changed files with 88 additions and 10 deletions

View File

@ -6636,15 +6636,18 @@ Show that $h$ is one-to-one.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.9}}% \subsection{\unverified{Exercise 4.9}}%
\hyperlabel{sub:exercise-4.9} \hyperlabel{sub:exercise-4.9}
Let $f$ be a function from $B$ into $B$, and assume that $A \subseteq B$. 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 We have two possible methods for constructing the "closure" $C$ of $A$ under
$f$. $f$.
First define $C^*$ to be the intersection of the closed supersets of $A$: First define $C^*$ to be the intersection of the closed supersets of $A$:
$$C^* = \bigcap\{X \mid \begin{equation}
A \subseteq X \subseteq B \land \img{f}{X} \subseteq X\}.$$ \label{sub:exercise-4.9-eq1}
C^* = \bigcap\{X \mid
A \subseteq X \subseteq B \land \img{f}{X} \subseteq X\}.
\end{equation}
Alternatively, we could apply the recursion theorem to obtain the function $h$ Alternatively, we could apply the recursion theorem to obtain the function $h$
for which for which
\begin{align*} \begin{align*}
@ -6652,8 +6655,12 @@ Alternatively, we could apply the recursion theorem to obtain the function $h$
h(n^+) & = h(n) \cup \img{f}{h(n)}. h(n^+) & = h(n) \cup \img{f}{h(n)}.
\end{align*} \end{align*}
Clearly $h(0) \subseteq h(1) \subseteq \cdots$; define $C_*$ to be Clearly $h(0) \subseteq h(1) \subseteq \cdots$; define $C_*$ to be
$\bigcup\ran{h}$; in other words $$C_* = \bigcup_{i \in \omega} h(i).$$ $\bigcup\ran{h}$; in other words
Show that $C^+ = C_*$. \begin{equation}
\label{sub:exercise-4.9-eq2}
C_* = \bigcup_{i \in \omega} h(i).
\end{equation}
Show that $C^* = C_*$.
[\textit{Suggestion}: [\textit{Suggestion}:
To show that $C^* \subseteq C_*$, show that $\img{f}{C_*} \subseteq C_*$. To show that $C^* \subseteq C_*$, show that $\img{f}{C_*} \subseteq C_*$.
To show that $C_* \subseteq C^*$, use induction to show that To show that $C_* \subseteq C^*$, use induction to show that
@ -6661,11 +6668,70 @@ To show that $C_* \subseteq C^*$, use induction to show that
\begin{proof} \begin{proof}
TODO We show that $C^* \subseteq C_*$ and $C^* \supseteq C_*$.
\paragraph{($\subseteq$)}%
It suffices to show $\img{f}{C_*} \subseteq C_*$ since then $C_*$ is a
member of the family of sets being intersected in
\eqref{sub:exercise-4.9-eq1}.
Let $t \in \img{f}{C_*}$.
By definition of the \nameref{ref:image} of a set, there exists some
$u \in C_*$ such that $f(u) = t$.
By \eqref{sub:exercise-4.9-eq2}, there exists some $i \in \omega$ such that
$u \in h(i)$.
Then $t \in \img{f}{h(i)} \subseteq h(i) \cup \img{f}{h(i)} = h(i^+)$.
Therefore \eqref{sub:exercise-4.9-eq2} indicates $t \in C_*$.
\paragraph{($\supseteq$)}%
Define $$S = \{n \in \omega \mid h(n) \subseteq C^*\}.$$
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterward we prove that (iii) $C_* \subseteq C^*$.
\subparagraph{(i)}%
\label{spar:exercise-4.9-i}
By construction, $h(0) = A$.
It trivially follows that $A \subseteq C^*$ by
\eqref{sub:exercise-4.9-eq1}.
Thus $h(0) \subseteq C^*$ meaning $0 \in S$.
\subparagraph{(ii)}%
\label{spar:exercise-4.9-ii}
Suppose $n \in S$.
That is, $h(n) \subseteq C^*$.
We must prove that $h(n^+) \subseteq C^*$.
Let $$t \in h(n^+) = h(n) \cup \img{f}{h(n)}.$$
Then either $t \in h(n)$ or $t \in \img{f}{h(n)}$.
If $t \in h(n)$, it immediately follows $t \in C^*$.
If $t \in \img{f}{h(n)}$, then the definition of the \nameref{ref:image}
of a set implies there exists some $u \in h(n)$ such that $f(u) = t$.
Since $h(n) \subseteq C^*$, \eqref{sub:exercise-4.9-eq1} indicates that
$$\forall X, A \subseteq X \subseteq B \land
\img{f}{X} \subseteq X \Rightarrow u \in X.$$
But then closure under image yields
$$\forall X, A \subseteq X \subseteq B \land
\img{f}{X} \subseteq X \Rightarrow f(u) \in X.$$
Since $f(u) = t$, $t \in C^*$.
Thus $h(n^+) \subseteq C^*$, i.e. $n^+ \in S$.
\subparagraph{(iii)}%
By \nameref{spar:exercise-4.9-i} and \nameref{spar:exercise-4.9-ii},
$S \subseteq \omega$ is an \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
That is, for all $n \in \omega$, $h(n) \subseteq C^*$.
Thus $$C_* = \bigcup_{i \in \omega} h(i) \subseteq C^*.$$
\paragraph{Conclusion}%
Since $C^* \subseteq C_*$ and $C_* \subseteq C^*$, it follows $C^* = C_*$.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.10}}% \subsection{\unverified{Exercise 4.10}}%
\hyperlabel{sub: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 $A$ In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x^2$, and $A$
@ -6674,11 +6740,17 @@ What is the set called $C^*$ and $C_*$?
\begin{proof} \begin{proof}
TODO By \nameref{sub:exercise-4.9}, $C^* = C_*$.
By definition,
$$C^* = \bigcap \{X \mid
\icc{\frac{1}{2}}{1} \subseteq X \subseteq \mathbb{R} \land
\img{f}{X} \subseteq X\}.$$
Since $f(x)$ converges to $0$ for all values $x < 1$, it follows
$C^* = C_* = \ioc{0}{1}$.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.11}}% \subsection{\unverified{Exercise 4.11}}%
\hyperlabel{sub: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 In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x - 1$, and
@ -6687,7 +6759,13 @@ What is the set called $C^*$ and $C_*$?
\begin{proof} \begin{proof}
TODO By \nameref{sub:exercise-4.9}, $C^* = C_*$.
By definition, $$C_* = \bigcup_{i \in \omega} h(i)$$ where
\begin{align*}
h(0) & = A = \{0\}, \\
h(n^+) & = h(n) \cup \img{f}{h(n)}.
\end{align*}
Thus $C_* = C^* = \{x \in \mathbb{Z} \mid x \leq 0\}$.
\end{proof} \end{proof}