From 6eaea4b6a0fde9ca174460b40246d8c06adf42bf Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Tue, 22 Aug 2023 08:01:33 -0600 Subject: [PATCH] Endreton (logic). Proofs of "Finite Set" theorems. --- Bookshelf/Enderton/Set.tex | 237 +++++++++++++++++++++++++++++++++++-- 1 file changed, 224 insertions(+), 13 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index f327f97..728cd53 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -8507,7 +8507,7 @@ \end{proof} -\subsection{\unverified{Theorem 6B}}% +\subsection{\pending{Theorem 6B}}% \hyperlabel{sub:theorem-6b} \begin{theorem}[6B] @@ -8530,7 +8530,7 @@ \section{Finite Sets}% \hyperlabel{sec:finite-sets} -\subsection{\sorry{Pigeonhole Principle}}% +\subsection{\pending{Pigeonhole Principle}}% \hyperlabel{sub:pigeonhole-principle} \begin{theorem} @@ -8538,10 +8538,70 @@ \end{theorem} \begin{proof} - TODO + + Let + \begin{equation} + \hyperlabel{sub:pigeonhole-principle-eq1} + S = \{n \in \omega \mid + \forall m \in n, \text{every one-to-one function } + f \colon m \rightarrow n \text{ is not onto}\}. + \end{equation} + We show that (i) $0 \in S$ and (ii) if $n \in S$, then so is $n^+$. + Afterward we prove (iii) the theorem statement. + + \paragraph{(i)}% + \hyperlabel{par:pigeonhole-principle-i} + + By \nameref{sub:zero-least-natural-number}, $0$ is the least natural + number. + Therefore, $0 \in S$ vacuously. + + \paragraph{(ii)}% + \hyperlabel{par:pigeonhole-principle-ii} + + Suppose $n \in S$ and let $m \in n^+$. + Furthermore, let $f \colon m \rightarrow n^+$ be a one-to-one + \nameref{ref:function} (as proof of a one-to-one function's existence, + just consider the identity function). + There are two cases to consider: + + \subparagraph{Case 1}% + + Suppose $n^+ \not\in \ran{f}$. + Then it obviously follows $f$ is not onto $n^+$. + + \subparagraph{Case 2}% + + Suppose $n^+ \in \ran{f}$. + Then there exists some $t$ such that $\tuple{t, n^+} \in f$. + Consider $f' = f \restriction n$. + Since $f$ is one-to-one, it follows $f'$ is also one-to-one. + By \eqref{sub:pigeonhole-principle-eq1}, $f'$ is not onto $n$. + That is, there exists some $p \in n$ such that $p \not\in \ran{f'}$. + But since $p \in n \in n^+$, \nameref{sub:theorem-4f} implies + $p \in n^+$. + Furthermore, $p \not\in \ran{f}$ by virtue of how $f'$ was constructed. + Thus $f$ is not onto $n^+$. + + \subparagraph{Subconclusion}% + + The foregoing cases are exhaustive. + Hence $n^+ \in S$. + + \paragraph{(iii)}% + + By \nameref{par:pigeonhole-principle-i} and + \nameref{par:pigeonhole-principle-ii}, $S$ is an + \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Thus for all natural numbers $n$, there is no one-to-one correspondence + between $n$ and a proper subset of $n$. + In other words, no natural number is equinumerous to a proper subset of + itself. + \end{proof} -\subsection{\sorry{Corollary 6C}}% +\subsection{\pending{Corollary 6C}}% \hyperlabel{sub:corollary-6c} \begin{corollary}[6C] @@ -8549,13 +8609,31 @@ \end{corollary} \begin{proof} - TODO + Let $S$ be a \nameref{ref:finite-set} and $S'$ be a + \nameref{ref:proper-subset} $S'$ of $S$. + Then there exists some nonempty set $T$, disjoint from $S'$, such that + $S' \cup T = S$. + By definition of a finite set, $S$ is \nameref{ref:equinumerous} to a + natural number $n$. + By \nameref{sub:theorem-6a}, $\equinumerous{S' \cup T}{S}$ which, by the + same theorem, implies $\equinumerous{S' \cup T}{n}$. + + Let $f$ be a one-to-one correspondence between $S' \cup T$ and $n$. + Since $T$ is nonempty, $f \restriction S'$ is a one-to-one correspondence + between $S'$ and a proper subset of $n$. + By the \nameref{sub:pigeonhole-principle}, $n$ is not equinumerous to any + proper subset of itself. + Therefore \nameref{sub:theorem-6a} implies $S'$ cannot be equinumerous to + $n$, which, by the same theorem, implies $S'$ cannot be equinumerous to + $S$. + Hence no finite set is equinumerous to a proper subset of itself. \end{proof} -\subsection{\sorry{Corollary 6D}}% +\subsection{\pending{Corollary 6D}}% \hyperlabel{sub:corollary-6d} \begin{corollary}[6D] + \ % Force a newline. \begin{enumerate}[(a)] \item Any set equinumerous to a proper subset of itself is infinite. \item The set $\omega$ is infinite. @@ -8563,10 +8641,56 @@ \end{corollary} \begin{proof} - TODO + + \paragraph{(a)}% + \hyperlabel{par:corollary-6d-a} + + Let $S$ be a set \nameref{ref:equinumerous} to \nameref{ref:proper-subset} + $S'$ of itself. + Then $S$ cannot be a \nameref{ref:finite-set} by + \nameref{sub:corollary-6c}. + By definition, $S$ is an \nameref{ref:infinite-set}. + + \paragraph{(b)}% + + Consider set $S = \{n \in \omega \mid n \text{ is even}\}$. + We prove that (i) $S$ is \nameref{ref:equinumerous} to $\omega$ and (ii) + that $\omega$ is infinite. + + \subparagraph{(i)}% + \hyperlabel{spar:corollary-6d-i} + + Define $f \colon \omega \rightarrow S$ given by $f(n) = 2 \cdot n$. + Notice $f$ is well-defined by the definition of an even natural number, + introduced in \nameref{sub:exercise-4.14}. + We first show $f$ is one-to-one and then that $f$ is onto. + + Suppose $f(n_1) = f(n_2) = 2 \cdot n_1$. + We must prove that $n_1 = n_2$. + By the \nameref{sub:trichotomy-law-natural-numbers}, exactly one of the + following may occur: $n_1 = n_2$, $n_1 < n_2$, or $n_2 < n_1$. + If $n_1 < n_2$, then \nameref{sub:theorem-4n} implies + $n_1 \cdot 2 < n_2 \cdot 2$. + \nameref{sub:theorem-4k-5} then indicates $2 \cdot n_1 < 2 \cdot n_2$, + a contradiction to $2 \cdot n_1 = 2 \cdot n_2$. + A parallel argument holds for when $n_2 < n_1$. + Thus $n_1 = n_2$. + + Next, let $m \in S$. + That is, $m$ is an even number. + By definition, there exists some $n \in \omega$ such that + $m = 2 \cdot n$. + Thus $f(n) = m$. + + \subparagraph{(ii)}% + + By \nameref{spar:corollary-6d-i}, $\omega$ is equinumerous to a subset + of itself. + By \nameref{par:corollary-6d-a}, $\omega$ is infinite. + \end{proof} -\subsection{\sorry{Corollary 6E}}% +\subsection{\pending{Corollary 6E}}% \hyperlabel{sub:corollary-6e} \begin{corollary}[6E] @@ -8574,10 +8698,22 @@ \end{corollary} \begin{proof} - TODO + Let $S$ be a \nameref{ref:finite-set}. + By definition $S$ is equinumerous to a natural number $n$. + Suppose $S$ is equinumerous to another natural number $m$. + By \nameref{sub:trichotomy-law-natural-numbers}, exactly one of three + situations is possible: $n = m$, $n < m$, or $m < n$. + + If $n < m$, then $\equinumerous{m}{S}$ and $\equinumerous{S}{n}$. + By \nameref{sub:theorem-6a}, it follows $\equinumerous{m}{n}$. + But \nameref{sub:pigeonhole-principle} indicates no natural number is + equinumerous to a proper subset of itself, a contradiction. + If $m < n$, a parallel argument applies. + Hence $n = m$, proving every finite set is equinumerous to a unique natural + number. \end{proof} -\subsection{\sorry{Lemma 6F}}% +\subsection{\pending{Lemma 6F}}% \hyperlabel{sub:lemma-6f} \begin{lemma}[6F] @@ -8586,10 +8722,73 @@ \end{lemma} \begin{proof} - TODO + + Let + \begin{equation} + \hyperlabel{sub:lemma-6f-eq1} + S = \{n \in \omega \mid \forall C \subset n, + \exists m < n \text{ such that } \equinumerous{C}{m}\}. + \end{equation} + We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. + Afterward we prove (iii) the lemma statement. + + \paragraph{(i)}% + \hyperlabel{par:lemma-6f-i} + + By definition, $0 = \emptyset$. + Thus $0$ has no proper subsets. + Hence $0 \in S$ vacuously. + + \paragraph{(ii)}% + \hyperlabel{par:lemma-6f-ii} + + Suppose $n \in S$ and consider $n^+$. + By definition of the \nameref{ref:successor}, $n^+ = n \cup \{n\}$. + Let $C$ be an arbitrary, \nameref{ref:proper-subset} of $n^+$. + There are two cases to consider: + + \subparagraph{Case 1}% + \hyperlabel{spar:lemma-6f-1} + + Suppose $n \not\in C$. + Then $C \subseteq n$. + If $C$ is a proper subset of $n$, \eqref{sub:lemma-6f-eq1} implies $C$ + is \nameref{ref:equinumerous} to some $m < n < n^+$. + If $C = n$, then \nameref{sub:theorem-6a} implies $C$ is equinumerous to + $n < n^+$. + + \subparagraph{Case 2}% + + Suppose $n \in C$. + Since $C$ is a proper subset of $n^+$, the set $n^+ - C$ is nonempty. + By \nameref{sub:well-ordering-natural-numbers}, $n^+ - C$ has a least + element, say $p$ (which does not equal $n$). + Consider now set $C' = (C - \{n\}) \cup \{p\}$. + By construction, $C' \subseteq n$. + As seen in \nameref{spar:lemma-6f-1}, $C'$ is equinumerous to some + $m < n^+$. + + It suffices to show there exists a one-to-one correspondence between + $C'$ and $C$, since then \nameref{sub:theorem-6a} implies $C$ is + equinumerous to $m$ as well. + Function $f \colon C' \rightarrow C$ given by + $$f(x) = \begin{cases} + n & \text{if } x = p \\ + x & \text{otherwise} + \end{cases}$$ + is trivially one-to-one and onto as expected. + + \paragraph{(iii)}% + + By \nameref{par:lemma-6f-i} and \nameref{par:lemma-6f-ii}, $S$ is an + \nameref{ref:inductive-set}. + By \nameref{sub:theorem-4b}, $S = \omega$. + Therefore, for every proper subset $C$ of a natural number $n$, there + exists some $m < n$ such that $\equinumerous{C}{n}$. + \end{proof} -\subsection{\sorry{Corollary 6G}}% +\subsection{\pending{Corollary 6G}}% \hyperlabel{sub:corollary-6g} \begin{corollary}[6G] @@ -8597,7 +8796,19 @@ \end{corollary} \begin{proof} - TODO + Let $S$ be a \nameref{ref:finite-set} and $S' \subseteq S$. + Clearly, if $S' = S$, then $S'$ is finite. + Therefore suppose $S'$ is a proper subset of $S$. + + By definition of finite set, $S$ is \nameref{ref:equinumerous} to some + natural number $n$. + Let $f$ be a one-to-one correspondence between $S$ and $n$. + Then $f \restriction S'$ is a one-to-one correspondence between $S'$ and + some proper subset of $n$. + By \nameref{sub:lemma-6f}, $\ran{(f \restriction S')}$ is equinumerous to + some $m < n$. + Then \nameref{sub:theorem-6a} indicates $\equinumerous{S'}{m}$. + Hence $S'$ is a finite set. \end{proof} \section{Exercises 6}%