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Enderton. Finish drafting prompts of Natural Numbers section.

finite-set-exercises
Joshua Potter 2023-08-05 11:18:53 -06:00
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Bookshelf/Enderton/Set.tex
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@ -7,6 +7,7 @@
\makeleancommands{../..}
\newcommand{\pair}[1]{\left< #1 \right>}
\newcommand{\ineq}{\,\mathop{\underline{\in}}\,}
\begin{document}
@ -446,7 +447,7 @@ Likewise, define $m$ to be \textbf{less than or equal to} $n$ if and only if
That is,
\begin{align*}
m \leq n
& \iff m \mathop{\underline{\in}} n \\
& \iff m \ineq n \\
& \iff m < n \lor m = n.
\end{align*}
@ -548,6 +549,19 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
\end{axiom}
\section{\defined{Proper Subset}}%
\hyperlabel{ref:proper-subset}
A set $A$ is said to be a \textbf{proper subset} of $B$ ($A \subset B$) if and
only if it is a subset of $B$ that is unequal to $B$.
$$A \subset B \iff A \subseteq B \land A \neq B.$$
\begin{definition}
\lean*{Std/Classes/SetNotation}{HasSSubset}
\end{definition}
\section{\defined{Quotient Set}}%
\hyperlabel{ref:quotient-set}
@ -7212,7 +7226,7 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\pending{Lemma 4L(a)}}%
\subsection{\verified{Lemma 4L(a)}}%
\hyperlabel{sub:lemma-4l-a}
\begin{lemma}[4L(a)]
@ -7221,6 +7235,10 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{lemma}
\begin{note}
Here I referred to Enderton's proof in the forward direction.
\end{note}
\begin{proof}
\lean{Std/Data/Nat/Lemmas}{Nat.succ\_lt\_succ\_iff}
@ -7229,13 +7247,50 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\paragraph{($\Rightarrow$)}%
Suppose $m \in n$.
TODO
Define $$S = \{n \in \omega \mid (\forall m \in n) m^+ \in n^+\}.$$
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterwards we show that (iii) the forward direction of the stated
biconditional holds.
\subparagraph{(i)}%
\hyperlabel{spar:lemma-4l-a-i}
$0 \in S$ vacuously.
That is, there are no members of $0 = \emptyset$ by definition.
\subparagraph{(ii)}%
\hyperlabel{spar:lemma-4l-a-ii}
Suppose $n \in S$.
We need to show for all $m \in n^+$, $m^+ \in n^{++}$.
Let $m \in n^+ = n \cup \{n\}$.
Then $m \in n$ or $m \in \{n\}$.
If $m \in n$, then $n \in S$ implies $m^+ \in n^+ \in n^{++}$.
By \nameref{sub:theorem-4f}, every natural number is a
\nameref{ref:transitive-set}.
Therefore $m^+ \in n^{++}$.
On the other hand, if $m \in \{n\}$, then $m = n$.
Since $n^+ \in n^{++}$, it immediately follows $m^+ \in n^{++}$.
Hence $n^+ \in S$.
\subparagraph{(iii)}%
By \nameref{spar:lemma-4l-a-i} and \nameref{spar:lemma-4l-a-ii}, $S$ is an
\nameref{ref:inductive-set}.
Hence \nameref{sub:theorem-4b} implies $S = \omega$.
Thus for all $n \in \omega$, $m \in n \Rightarrow m^+ \in n^+$.
\paragraph{($\Leftarrow$)}%
Suppose $m^+ \in n^+$.
TODO
The definition of \nameref{ref:successor} immediately implies that
$m \in m^+$.
By \nameref{sec:ordering-natural-numbers}, $m^+ \in n^+$ implies $m^+ \in n$
or $m^+ = n$.
If the latter, $m \in n$ immediately follows.
If the former, we note $n$ is a transitive set by \nameref{sub:theorem-4f}.
Thus $m \in m^+ \in n$ implies $m \in n$.
\end{proof}
@ -7449,6 +7504,166 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\end{proof}
\subsection{\unverified{Corollary 4M}}%
\hyperlabel{sub:corollary-4m}
\begin{corollary}[4M]
For any natural numbers $m$ and $n$,
\begin{equation}
\hyperlabel{sub:corollary-4m-eq1}
m \in n \iff m \subset n
\end{equation}
and
\begin{equation}
\hyperlabel{sub:corollary-4m-eq2}
m \ineq n \iff m \subseteq n.
\end{equation}
\end{corollary}
\begin{proof}
\paragraph{\eqref{sub:corollary-4m-eq1}}%
We prove both directions of the biconditional specified in
\eqref{sub:corollary-4m-eq1}:
\subparagraph{($\Rightarrow$)}%
Suppose $m \in n$ and $t \in m$.
By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}.
Therefore $t \in n$.
Hence $m \subseteq n$.
Since $m \in n$, \nameref{sub:linear-ordering-natural-numbers} implies
$m \neq n$.
Thus, by definition of \nameref{ref:proper-subset}, $m \subset n$.
\subparagraph{($\Leftarrow$)}%
Suppose $m \subset n$.
By \nameref{sub:linear-ordering-natural-numbers}, exactly one of
$$m \in n, \quad m = n, \quad n \in m$$ holds.
By definition of \nameref{ref:proper-subset}, $m \subseteq n$ and
$m \neq n$.
Furthermore, it cannot be that $n \in m$ since otherwise $n \in n$,
contradicting \nameref{sub:lemma-4l-b}.
Thus $m \in n$ is the only possibility.
\paragraph{\eqref{sub:corollary-4m-eq2}}%
We prove both directions of the biconditional specified in
\eqref{sub:corollary-4m-eq2}:
\subparagraph{($\Rightarrow$)}%
Suppose $m \ineq n$.
By definition, $m \in n$ or $m = n$.
Let $p \in m$.
Then $p \in m \in n$ or $p \in m = n$.
By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}.
Thus $p \in n$ in either case.
Hence $m \subseteq n$.
\subparagraph{($\Leftarrow$)}%
Suppose $m \subseteq n$.
By \nameref{sub:linear-ordering-natural-numbers}, exactly one of
$$m \in n, \quad m = n, \quad n \in m$$ holds.
But it cannot be that $n \in m$ since that would imply $n \in n$,
contradicting \nameref{sub:lemma-4l-b}.
Therefore $m \in n$ or $m = n$.
Hence $m \ineq n$.
\end{proof}
\subsection{\sorry{Theorem 4N}}%
\hyperlabel{sub:theorem-4n}
\begin{theorem}[4N]
For any natural numbers $n$, $m$, and $p$, $$m \in n \iff m + p \in n + p.$$
If, in addition, $p \neq 0$, then $$m \in n \iff m \cdot p \in n \cdot p.$$
\end{theorem}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Corollary 4P}}%
\hyperlabel{sub:corollary-4p}
\begin{corollary}[4P]
The following cancellation laws hold for $m$, $n$, and $p$ in $\omega$:
\begin{align*}
m + p = n + p & \Rightarrow m = n, \\
m \cdot p = n \cdot p & \Rightarrow m = n.
\end{align*}
\end{corollary}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{%
Well Ordering of \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:well-ordering-natural-numbers}
\begin{theorem}
Let $A$ be a nonempty subset of $\omega$.
Then there is some $m \in A$ such that $m \ineq n$ for all $n \in A$.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Corollary 4Q}}%
\hyperlabel{sub:corollary-4q}
\begin{corollary}[4Q]
There is no function $f \colon \omega \rightarrow \omega$ such that
$f(n^+) \in f(n)$ for every natural number $n$.
\end{corollary}
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{%
Strong Induction Principle for \texorpdfstring{$\omega$}{Natural Numbers}}}%
\hyperlabel{sub:strong-induction-principle-natural-numbers}
\begin{theorem}
Let $A$ be a subset of $\omega$, and assume that for every $n \in \omega$,
$$\text{if every number less than } n \text{ is in } A,
\text{ then } n \in A.$$
Then $A = \omega$.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\section{Exercises 4}%
\hyperlabel{sec:exercises-4}
@ -8027,4 +8242,141 @@ Prove that $m^{n+p} = m^n \cdot m^p$.
\end{proof}
\subsection{\sorry{Exercise 4.18}}%
\hyperlabel{sub:exercise-4.18}
Simplify $\img{\in_\omega^{-1}}{\{7, 8\}}$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.19}}%
\hyperlabel{sub:exercise-4.19}
Prove that if $m$ is a natural number and $d$ is a nonzero number, then there
exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r$ is less than
$d$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.20}}%
\hyperlabel{sub:exercise-4.20}
Let $A$ be a nonempty subset of $\omega$ such that $\bigcup A = A$.
Show that $A = \omega$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.21}}%
\hyperlabel{sub:exercise-4.21}
Show that no natural number is a subset of any of its elements.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.22}}%
\hyperlabel{sub:exercise-4.22}
Show that for any natural numbers $m$ and $p$ we have $m \in m + p^+$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.23}}%
\hyperlabel{sub:exercise-4.23}
Assume that $m$ and $n$ are natural numbers with $m$ less than $n$.
Show that there is some $p$ in $\omega$ for which $m + p^+ = n$.
(It follows form this and the preceding exercise that $m$ is less than $n$ iff
$(\exists p \in \omega)m + p^+ = n$.)
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.24}}%
\hyperlabel{sub:exercise-4.24}
Assume that $m + n = p + q$.
Show that $$m \in p \iff n \in q.$$
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.25}}%
\hyperlabel{sub:exercise-4.25}
Assume that $n \in m$ and $q \in p$.
Show that $$(m \cdot q) + (n \cdot p) \in (m \cdot p) + (n \cdot q).$$
[\textit{Suggestion:} Use \nameref{sub:exercise-4.23}.]
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.26}}%
\hyperlabel{sub:exercise-4.26}
Assume that $n \in \omega$ and $f \colon n^+ \rightarrow \omega$.
Show that $\ran{f}$ has a largest element.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.27}}%
\hyperlabel{sub:exercise-4.27}
Assume that $A$ is a set, $G$ is a function, and $f_1$ and $f_2$ map $\omega$
into $A$.
Further assume that for each $n \in \omega$ both $f_1 \restriction n$ and
$f_2 \restriction n$ belong to $\dom{G}$ and
$$f_1(n) = G(f_1 \restriction n) \land f_2(n) = G(f_2 \restriction n).$$
Show that $f_1 = f_2$.
\begin{proof}
TODO
\end{proof}
\subsection{\sorry{Exercise 4.28}}%
\hyperlabel{sub:exercise-4.28}
Rewrite the proof of \nameref{sub:theorem-4g} using, in place of induction, the
well-ordering of $\omega$.
\begin{proof}
TODO
\end{proof}
\end{document}

7
preamble.tex
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@ -99,6 +99,13 @@
\newenvironment{definition}{\@statement{Definition}}{\hfill$\square$}
\renewenvironment{proof}{\@statement{Proof}}{\hfill$\square$}
\newtheorem{corollaryinner}{Corollary}
\newenvironment{corollary}[1][]{%
\ifstrempty{#1}
{\corollaryinner}
{\renewcommand\thecorollaryinner{#1}\corollaryinner}
}{\endcorollaryinner}
\newtheorem{lemmainner}{Lemma}
\newenvironment{lemma}[1][]{%
\ifstrempty{#1}