Enderton (set). Theorem 4H.

finite-set-exercises
Joshua Potter 2023-08-27 13:48:10 -06:00
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\end{proof} \end{proof}
\subsection{\sorry{Theorem 4H}}% \subsection{\pending{Theorem 4H}}%
\hyperlabel{sub:theorem-4h} \hyperlabel{sub:theorem-4h}
\begin{theorem}[4H] \begin{theorem}[4H]
@ -6390,7 +6390,91 @@
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
TODO
Let $\langle N, S, e \rangle$ be a \nameref{ref:peano-system}.
By the \nameref{sub:recursion-theorem-natural-numbers}, there exists a
unique function $h \colon \omega \rightarrow N$ such that $h(0) = e$ and
for every $n \in \omega$, $h(n^+) = h(\sigma(n)) = S(h(n))$.
All that remains is proving $h$ is one-to-one and onto.
\suitdivider
\noindent
We first show $h$ is one-to-one by induction.
Define
$$S = \{n \in \omega \mid \forall m, h(m) = h(n) \Rightarrow m = n\}.$$
We show that (i) $0 \in S$ and (ii) if $n \in S$, then $n^+ \in S$.
Afterward we show (iii) that $h$ is one-to-one.
\paragraph{(i)}%
\hyperlabel{par:theorem-4h-i}
Let $m \in \omega$ such that $h(m) = h(0) = e$.
Suppose $m \neq 0$.
Then \nameref{sub:theorem-4c} indicates there exists some natural number
$p$ such that $p^+ = m$.
But then $h(m) = h(p^+) = S(h(p)) = e$.
By definition of a Peano system, $e \not\in \ran{S}$.
This is a contradiction.
Hence $m = 0$, i.e. $0 \in S$.
\paragraph{(ii)}%
\hyperlabel{par:theorem-4h-ii}
Suppose $n \in S$.
Let $m \in \omega$ such that $h(m) = h(n^+) = S(h(n))$.
There are two cases to consider:
\subparagraph{Case 1}%
Suppose $m = 0$.
Then $h(0) = e = S(h(n))$.
But, by definition of a Peano system, $e \not\in \ran{S}$.
Thus we have a contradiction.
\subparagraph{Case 2}%
Suppose $m \neq 0$.
Then \nameref{sub:theorem-4c} indicates there exists some natural number
$p$ such that $p^+ = m$.
Then $$S(h(n)) = h(m) = h(p^+) = S(h(p)).$$
By definition of a Peano system, $S$ is one-to-one.
Therefore $h(n) = h(p)$.
Since $n \in S$, it follows $n = p$.
Therefore $n^+ = p^+ = m$.
\subparagraph{Subconclusion}%
The above two cases are exhaustive.
Hence $n^+ \in S$.
\paragraph{(iii)}%
By \nameref{par:theorem-4h-i} and \nameref{par:theorem-4h-ii}, $S$ is an
\nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$.
Thus for all natural numbers $m, n \in \omega$, if $h(m) = h(n)$, then
$m = n$.
In other words, $f$ is one-to-one.
\suitdivider
\noindent
We next show that $\ran{h} = N$.
By the Peano posulate, every subset $A$ of $N$ containing $e$ and closed under $S$ is $N$ itself.
Thus it suffices to prove that (i) $e \in \ran{h}$ and (ii) $\ran{h}$ is closed under $S$.
\paragraph{(i)}%
That is, $h(0) = e$ by definition.
Thus $e \in \ran{h}$.
\paragraph{(ii)}%
Let $y \in \ran{h}$.
Then there exists some $n \in \omega$ such that $h(n) = y$.
By definition of $h$, $h(n^+) = S(h(n)) = S(y)$.
Therefore $S(y) \in \ran{h}$.
Since this holds for any $y \in \ran{h}$, $\ran{h}$ is closed under $S$.
\end{proof} \end{proof}
\section{Arithmetic}% \section{Arithmetic}%