Endreton (logic). Proofs of "Finite Set" theorems.
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\end{proof}
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\subsection{\unverified{Theorem 6B}}%
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\subsection{\pending{Theorem 6B}}%
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\hyperlabel{sub:theorem-6b}
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\begin{theorem}[6B]
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@ -8530,7 +8530,7 @@
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\section{Finite Sets}%
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\hyperlabel{sec:finite-sets}
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\subsection{\sorry{Pigeonhole Principle}}%
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\subsection{\pending{Pigeonhole Principle}}%
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\hyperlabel{sub:pigeonhole-principle}
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\begin{theorem}
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@ -8538,10 +8538,70 @@
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\end{theorem}
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\begin{proof}
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TODO
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Let
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\begin{equation}
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\hyperlabel{sub:pigeonhole-principle-eq1}
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S = \{n \in \omega \mid
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\forall m \in n, \text{every one-to-one function }
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f \colon m \rightarrow n \text{ is not onto}\}.
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\end{equation}
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We show that (i) $0 \in S$ and (ii) if $n \in S$, then so is $n^+$.
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Afterward we prove (iii) the theorem statement.
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\paragraph{(i)}%
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\hyperlabel{par:pigeonhole-principle-i}
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By \nameref{sub:zero-least-natural-number}, $0$ is the least natural
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number.
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Therefore, $0 \in S$ vacuously.
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\paragraph{(ii)}%
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\hyperlabel{par:pigeonhole-principle-ii}
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Suppose $n \in S$ and let $m \in n^+$.
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Furthermore, let $f \colon m \rightarrow n^+$ be a one-to-one
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\nameref{ref:function} (as proof of a one-to-one function's existence,
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just consider the identity function).
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $n^+ \not\in \ran{f}$.
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Then it obviously follows $f$ is not onto $n^+$.
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\subparagraph{Case 2}%
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Suppose $n^+ \in \ran{f}$.
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Then there exists some $t$ such that $\tuple{t, n^+} \in f$.
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Consider $f' = f \restriction n$.
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Since $f$ is one-to-one, it follows $f'$ is also one-to-one.
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By \eqref{sub:pigeonhole-principle-eq1}, $f'$ is not onto $n$.
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That is, there exists some $p \in n$ such that $p \not\in \ran{f'}$.
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But since $p \in n \in n^+$, \nameref{sub:theorem-4f} implies
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$p \in n^+$.
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Furthermore, $p \not\in \ran{f}$ by virtue of how $f'$ was constructed.
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Thus $f$ is not onto $n^+$.
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\subparagraph{Subconclusion}%
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The foregoing cases are exhaustive.
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Hence $n^+ \in S$.
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\paragraph{(iii)}%
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By \nameref{par:pigeonhole-principle-i} and
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\nameref{par:pigeonhole-principle-ii}, $S$ is an
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\nameref{ref:inductive-set}.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Thus for all natural numbers $n$, there is no one-to-one correspondence
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between $n$ and a proper subset of $n$.
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In other words, no natural number is equinumerous to a proper subset of
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itself.
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\end{proof}
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\subsection{\sorry{Corollary 6C}}%
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\subsection{\pending{Corollary 6C}}%
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\hyperlabel{sub:corollary-6c}
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\begin{corollary}[6C]
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@ -8549,13 +8609,31 @@
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\end{corollary}
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\begin{proof}
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TODO
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Let $S$ be a \nameref{ref:finite-set} and $S'$ be a
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\nameref{ref:proper-subset} $S'$ of $S$.
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Then there exists some nonempty set $T$, disjoint from $S'$, such that
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$S' \cup T = S$.
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By definition of a finite set, $S$ is \nameref{ref:equinumerous} to a
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natural number $n$.
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By \nameref{sub:theorem-6a}, $\equinumerous{S' \cup T}{S}$ which, by the
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same theorem, implies $\equinumerous{S' \cup T}{n}$.
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Let $f$ be a one-to-one correspondence between $S' \cup T$ and $n$.
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Since $T$ is nonempty, $f \restriction S'$ is a one-to-one correspondence
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between $S'$ and a proper subset of $n$.
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By the \nameref{sub:pigeonhole-principle}, $n$ is not equinumerous to any
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proper subset of itself.
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Therefore \nameref{sub:theorem-6a} implies $S'$ cannot be equinumerous to
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$n$, which, by the same theorem, implies $S'$ cannot be equinumerous to
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$S$.
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Hence no finite set is equinumerous to a proper subset of itself.
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\end{proof}
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\subsection{\sorry{Corollary 6D}}%
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\subsection{\pending{Corollary 6D}}%
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\hyperlabel{sub:corollary-6d}
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\begin{corollary}[6D]
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\ % Force a newline.
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\begin{enumerate}[(a)]
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\item Any set equinumerous to a proper subset of itself is infinite.
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\item The set $\omega$ is infinite.
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@ -8563,10 +8641,56 @@
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\end{corollary}
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\begin{proof}
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TODO
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\paragraph{(a)}%
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\hyperlabel{par:corollary-6d-a}
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Let $S$ be a set \nameref{ref:equinumerous} to \nameref{ref:proper-subset}
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$S'$ of itself.
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Then $S$ cannot be a \nameref{ref:finite-set} by
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\nameref{sub:corollary-6c}.
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By definition, $S$ is an \nameref{ref:infinite-set}.
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\paragraph{(b)}%
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Consider set $S = \{n \in \omega \mid n \text{ is even}\}$.
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We prove that (i) $S$ is \nameref{ref:equinumerous} to $\omega$ and (ii)
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that $\omega$ is infinite.
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\subparagraph{(i)}%
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\hyperlabel{spar:corollary-6d-i}
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Define $f \colon \omega \rightarrow S$ given by $f(n) = 2 \cdot n$.
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Notice $f$ is well-defined by the definition of an even natural number,
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introduced in \nameref{sub:exercise-4.14}.
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We first show $f$ is one-to-one and then that $f$ is onto.
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Suppose $f(n_1) = f(n_2) = 2 \cdot n_1$.
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We must prove that $n_1 = n_2$.
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By the \nameref{sub:trichotomy-law-natural-numbers}, exactly one of the
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following may occur: $n_1 = n_2$, $n_1 < n_2$, or $n_2 < n_1$.
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If $n_1 < n_2$, then \nameref{sub:theorem-4n} implies
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$n_1 \cdot 2 < n_2 \cdot 2$.
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\nameref{sub:theorem-4k-5} then indicates $2 \cdot n_1 < 2 \cdot n_2$,
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a contradiction to $2 \cdot n_1 = 2 \cdot n_2$.
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A parallel argument holds for when $n_2 < n_1$.
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Thus $n_1 = n_2$.
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Next, let $m \in S$.
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That is, $m$ is an even number.
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By definition, there exists some $n \in \omega$ such that
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$m = 2 \cdot n$.
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Thus $f(n) = m$.
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\subparagraph{(ii)}%
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By \nameref{spar:corollary-6d-i}, $\omega$ is equinumerous to a subset
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of itself.
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By \nameref{par:corollary-6d-a}, $\omega$ is infinite.
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\end{proof}
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\subsection{\sorry{Corollary 6E}}%
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\subsection{\pending{Corollary 6E}}%
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\hyperlabel{sub:corollary-6e}
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\begin{corollary}[6E]
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\end{corollary}
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\begin{proof}
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TODO
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Let $S$ be a \nameref{ref:finite-set}.
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By definition $S$ is equinumerous to a natural number $n$.
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Suppose $S$ is equinumerous to another natural number $m$.
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By \nameref{sub:trichotomy-law-natural-numbers}, exactly one of three
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situations is possible: $n = m$, $n < m$, or $m < n$.
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If $n < m$, then $\equinumerous{m}{S}$ and $\equinumerous{S}{n}$.
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By \nameref{sub:theorem-6a}, it follows $\equinumerous{m}{n}$.
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But \nameref{sub:pigeonhole-principle} indicates no natural number is
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equinumerous to a proper subset of itself, a contradiction.
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If $m < n$, a parallel argument applies.
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Hence $n = m$, proving every finite set is equinumerous to a unique natural
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number.
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\end{proof}
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\subsection{\sorry{Lemma 6F}}%
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\subsection{\pending{Lemma 6F}}%
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\hyperlabel{sub:lemma-6f}
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\begin{lemma}[6F]
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@ -8586,10 +8722,73 @@
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\end{lemma}
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\begin{proof}
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TODO
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Let
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\begin{equation}
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\hyperlabel{sub:lemma-6f-eq1}
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S = \{n \in \omega \mid \forall C \subset n,
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\exists m < n \text{ such that } \equinumerous{C}{m}\}.
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\end{equation}
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We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
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Afterward we prove (iii) the lemma statement.
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\paragraph{(i)}%
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\hyperlabel{par:lemma-6f-i}
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By definition, $0 = \emptyset$.
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Thus $0$ has no proper subsets.
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Hence $0 \in S$ vacuously.
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\paragraph{(ii)}%
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\hyperlabel{par:lemma-6f-ii}
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Suppose $n \in S$ and consider $n^+$.
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By definition of the \nameref{ref:successor}, $n^+ = n \cup \{n\}$.
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Let $C$ be an arbitrary, \nameref{ref:proper-subset} of $n^+$.
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There are two cases to consider:
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\subparagraph{Case 1}%
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\hyperlabel{spar:lemma-6f-1}
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Suppose $n \not\in C$.
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Then $C \subseteq n$.
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If $C$ is a proper subset of $n$, \eqref{sub:lemma-6f-eq1} implies $C$
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is \nameref{ref:equinumerous} to some $m < n < n^+$.
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If $C = n$, then \nameref{sub:theorem-6a} implies $C$ is equinumerous to
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$n < n^+$.
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\subparagraph{Case 2}%
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Suppose $n \in C$.
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Since $C$ is a proper subset of $n^+$, the set $n^+ - C$ is nonempty.
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By \nameref{sub:well-ordering-natural-numbers}, $n^+ - C$ has a least
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element, say $p$ (which does not equal $n$).
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Consider now set $C' = (C - \{n\}) \cup \{p\}$.
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By construction, $C' \subseteq n$.
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As seen in \nameref{spar:lemma-6f-1}, $C'$ is equinumerous to some
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$m < n^+$.
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It suffices to show there exists a one-to-one correspondence between
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$C'$ and $C$, since then \nameref{sub:theorem-6a} implies $C$ is
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equinumerous to $m$ as well.
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Function $f \colon C' \rightarrow C$ given by
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$$f(x) = \begin{cases}
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n & \text{if } x = p \\
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x & \text{otherwise}
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\end{cases}$$
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is trivially one-to-one and onto as expected.
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\paragraph{(iii)}%
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By \nameref{par:lemma-6f-i} and \nameref{par:lemma-6f-ii}, $S$ is an
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\nameref{ref:inductive-set}.
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By \nameref{sub:theorem-4b}, $S = \omega$.
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Therefore, for every proper subset $C$ of a natural number $n$, there
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exists some $m < n$ such that $\equinumerous{C}{n}$.
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\end{proof}
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\subsection{\sorry{Corollary 6G}}%
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\subsection{\pending{Corollary 6G}}%
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\hyperlabel{sub:corollary-6g}
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\begin{corollary}[6G]
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@ -8597,7 +8796,19 @@
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\end{corollary}
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\begin{proof}
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TODO
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Let $S$ be a \nameref{ref:finite-set} and $S' \subseteq S$.
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Clearly, if $S' = S$, then $S'$ is finite.
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Therefore suppose $S'$ is a proper subset of $S$.
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By definition of finite set, $S$ is \nameref{ref:equinumerous} to some
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natural number $n$.
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Let $f$ be a one-to-one correspondence between $S$ and $n$.
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Then $f \restriction S'$ is a one-to-one correspondence between $S'$ and
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some proper subset of $n$.
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By \nameref{sub:lemma-6f}, $\ran{(f \restriction S')}$ is equinumerous to
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some $m < n$.
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Then \nameref{sub:theorem-6a} indicates $\equinumerous{S'}{m}$.
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Hence $S'$ is a finite set.
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\end{proof}
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\section{Exercises 6}%
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