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Enderton. Part of ordering theorems/exercises.

finite-set-exercises
Joshua Potter 2023-08-11 07:00:57 -06:00
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Bookshelf/Enderton/Set.tex
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@ -7304,7 +7304,7 @@
\end{proof} \end{proof}
\subsection{\sorry{Corollary 4Q}}% \subsection{\unverified{Corollary 4Q}}%
\hyperlabel{sub:corollary-4q} \hyperlabel{sub:corollary-4q}
\begin{corollary}[4Q] \begin{corollary}[4Q]
@ -7313,7 +7313,11 @@
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
TODO Let $f \colon \omega \rightarrow \omega$.
Then $\emptyset \subset \ran{f} \subseteq \omega$.
By the \nameref{sub:well-ordering-natural-numbers}, $\ran{f}$ must have
a least element.
Therefore it isn't possible $f(n^+) \in f(n)$ for all $n \in \omega$.
\end{proof} \end{proof}
\subsection{\sorry{% \subsection{\sorry{%
@ -7322,13 +7326,22 @@
\begin{theorem} \begin{theorem}
Let $A$ be a subset of $\omega$, and assume that for every $n \in \omega$, Let $A$ be a subset of $\omega$, and assume that for every $n \in \omega$,
$$\text{if every number less than } n \text{ is in } A, \begin{equation}
\text{ then } n \in A.$$ \hyperlabel{sub:strong-induction-principle-natural-numbers-eq1}
\text{if every number less than } n \text{ is in } A,
\text{ then } n \in A.
\end{equation}
Then $A = \omega$. Then $A = \omega$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
TODO For the sake of contradiction, suppose $\omega - A$ is a nonempty set.
By \nameref{sub:well-ordering-natural-numbers}, there exists a least element
$m \in \omega - A$.
Then every number less than $m$ is in $A$.
But then \eqref{sub:strong-induction-principle-natural-numbers-eq1} implies
$m \in A$, a contradiction.
Thus $\omega - A$ is an empty set meaning $A = \omega$.
\end{proof} \end{proof}
\section{Exercises 4}% \section{Exercises 4}%
@ -7889,13 +7902,18 @@
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.18}}% \subsection{\unverified{Exercise 4.18}}%
\hyperlabel{sub:exercise-4.18} \hyperlabel{sub:exercise-4.18}
Simplify $\img{\in_\omega^{-1}}{\{7, 8\}}$. Simplify $\img{\in_\omega^{-1}}{\{7, 8\}}$.
\begin{proof} \begin{proof}
TODO By definition,
$$\in_\omega = \{\pair{m, n} \in \omega \times \omega \mid m \in n\}.$$
Thus, the \nameref{ref:inverse} of $\in_\omega$ is
$$\in_\omega^{-1} =
\{\pair{n, m} \in \omega \times \omega \mid m \in n\}.$$
Therefore $$\img{\in_\omega^{-1}}{\{7, 8\}} = \{6, 7\}.$$
\end{proof} \end{proof}
\subsection{\sorry{Exercise 4.19}}% \subsection{\sorry{Exercise 4.19}}%