From 76d294a47f66fe53fcdb8898527d39b40eb767d1 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Wed, 28 Jun 2023 13:19:59 -0600 Subject: [PATCH] Enderton. Finish function theorems. --- Bookshelf/Enderton/Set.tex | 237 ++++++++++++++++++++++++++++++++----- preamble.tex | 1 + 2 files changed, 207 insertions(+), 31 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index ef6d7f1..6c2203a 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -9,6 +9,7 @@ \newcommand{\dom}[1]{\textop{dom}{#1}} \newcommand{\fld}[1]{\textop{fld}{#1}} \newcommand{\ran}[1]{\textop{ran}{#1}} +\newcommand{\img}[2]{#1\left\llbracket#2\right\rrbracket} \begin{document} @@ -125,7 +126,7 @@ One-to-one functions are sometimes called \textbf{injections}. Let $A$ and $F$ be arbitrary sets. The \textbf{image of $A$ under $F$} is the set \begin{align*} - F[A] + \img{F}{A} & = \ran{(F \restriction A)} \\ & = \{v \mid (\exists u \in A) uFv\}. \end{align*} @@ -3413,59 +3414,233 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \end{proof} -\subsection{\unverified{Theorem 3K}}% -\label{sub:theorem-3k} +\subsection{\partial{Theorem 3K(a)}}% +\label{sub:theorem-3k-a} -\begin{theorem}[3K] +\begin{theorem}[3K(a)] The following hold for any sets. ($F$ need not be a function.) - - \begin{enumerate}[(a)] - \item The image of a union is the union of the images: - $$F[A \cup B] = F[A] \cup F[B] \quad\text{and}\quad - F\left[\bigcup{\mathscr{A}}\right] = - \bigcup\;\{F[A] \mid A \in \mathscr{A}\}.$$ - \item The image of an intersection is included in the intersection of the - images: - $$F[A \cap B] \subseteq F[A] \cap F[B] \quad\text{and}\quad - F\left[\bigcap\mathscr{A}\right] \subseteq - \bigcap\;\{F[A] \mid A \in \mathscr{A}\}$$ - for nonempty $\mathscr{A}$. - Equality holds if $F$ is single-rooted. - \item The image of a difference includes the difference of the images: - $$F[A] - F[B] \subseteq F[A - B].$$ - Equality holds if $F$ is single-rooted. - \end{enumerate} + The image of a union is the union of the images: + \begin{equation} + \label{sub:theorem-3k-a-eq1} + \img{F}{A \cup B} = \img{F}{A} \cup \img{F}{B} + \end{equation} + and + \begin{equation} + \label{sub:theorem-3k-a-eq2} + \img{F}{\bigcup{\mathscr{A}}} = + \bigcup\;\{\img{F}{A} \mid A \in \mathscr{A}\}. + \end{equation} \end{theorem} \begin{proof} - TODO + Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets. + We prove (i) \eqref{sub:theorem-3k-a-eq1} and (ii) + \eqref{sub:theorem-3k-a-eq2}. + + \paragraph{(i)}% + + By definition of the \nameref{ref:image} of a set: + \begin{align*} + \img{F}{A \cup B} + & = \{v \mid \exists u, u \in A \cup B \land uFv\} \\ + & = \{v \mid \exists u, + (u \in A \land uFv) \lor (u \in B \land uFv)\} \\ + & = \{v \mid (\exists u \in A) uFv\} \cup + \{v \mid (\exists u \in B) uFv\} \\ + & = \img{F}{A} \cup \img{F}{B}. + \end{align*} + + \paragraph{(ii)}% + + We prove that both sides of \eqref{sub:theorem-3k-a-eq2} is a subset of the + other. + + \subparagraph{($\subseteq$)}% + + Let $v \in \img{F}{\bigcup{\mathscr{A}}}$. + By definition of the \nameref{ref:image} of a set, there exists a set $u$ + such that $u \in \bigcup{\mathscr{A}} \land uFv$. + Then, by definition of the union of sets, there exists some + $A \in \mathscr{A}$ such that $u \in A$. + Therefore $v \in \img{F}{A}$ meaning + $v \in \bigcup\{\img{F}{A} \mid A \in \mathscr{A}\}$. + + \subparagraph{($\supseteq$)}% + + Let $v \in \bigcup\{\img{F}{A} \mid A \in \mathscr{A}\}$. + Then there exists some $b \in \{\img{F}{A} \mid A \in \mathscr{A}\}$ such + that $v \in b$. + In other words, there exists some $A \in \mathscr{A}$ such that + $v \in b = \img{F}{A}$. + By definition of the \nameref{ref:image} of a set, there exists a set $u$ + such that $u \in A \land uFv$. + But this implies that $u \in \bigcup{\mathscr{A}} \land uFv$. + Therefore $v \in \img{F}{\bigcup{\mathscr{A}}}$. \end{proof} -\subsection{\unverified{Corollary 3L}}% +\subsection{\partial{Theorem 3K(b)}}% +\label{sub:theorem-3k-b} + +\begin{theorem}[3K(b)] + + The following hold for any sets. ($F$ need not be a function.) + The image of an intersection is included in the intersection of the images: + \begin{equation} + \label{sub:theorem-3k-b-eq1} + \img{F}{A \cap B} \subseteq \img{F}{A} \cap \img{F}{B} + \end{equation} + and + \begin{equation} + \label{sub:theorem-3k-b-eq2} + \img{F}{\bigcap\mathscr{A}} \subseteq + \bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}. + \end{equation} + Equality holds if $F$ is single-rooted. + +\end{theorem} + +\begin{proof} + + Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets. + We first prove (i) \eqref{sub:theorem-3k-b-eq1} and (ii) + \eqref{sub:theorem-3k-b-eq2}. + Then, assuming $F$ is single-rooted, we prove both (iii) + \eqref{sub:theorem-3k-b-eq1} and (iv) \eqref{sub:theorem-3k-b-eq2} hold + under equality. + + \paragraph{(i)}% + \label{par:theorem-3k-b-i} + + Let $v \in \img{F}{A \cap B}$. + By definition of the \nameref{ref:image} of a set, + $\exists u \in A \cap B, uFv$. + Then $u \in A \land uFv$ and $u \in B \land uFv$. + Therefore $v \in \img{F}{A} \cap \img{F}{B}$. + + \paragraph{(ii)}% + \label{par:theorem-3k-b-ii} + + Let $v \in \img{F}{\bigcap{\mathscr{A}}}$. + By definition of the \nameref{ref:image} of a set, + $\exists u \in \bigcap{\mathscr{A}}, uFv$. + Then $\exists u, (\forall A \in \mathscr{A}, u \in A) \land uFv$. + This implies that $\forall A \in \mathscr{A}, \exists u \in A, uFv$. + Then $\forall A \in \mathscr{A}, v \in \img{F}{A}$. + Thus $v \in \bigcap\{\img{F}{A} \mid A \in \mathscr{A}\}$. + + \paragraph{(iii)}% + + Suppose $F$ is single-rooted. + By \nameref{par:theorem-3k-b-i}, + $$\img{F}{A \cap B} \subseteq \img{F}{A} \cap \img{F}{B}.$$ + All that remains is showing + $$\img{F}{A} \cap \img{F}{B} \subseteq \img{F}{A \cap B}.$$ + Let $v \in \img{F}{A} \cap \img{F}{B}$. + Then $v \in \img{F}{A}$ and $v \in \img{F}{B}$. + That is, $\exists u \in A, uFv$ and $\exists w \in B, wFv$. + Since $F$ is single rooted, it follows $u = w$. + Thus $u \in A \cap B \land uFv$ meaning $v \in \img{F}{A \cap B}$. + + \paragraph{(iv)}% + + Suppose $F$ is single-rooted. + By \nameref{par:theorem-3k-b-ii}, + $$\img{F}{\bigcap\mathscr{A}} \subseteq + \bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}.$$ + All that remains is showing + $$\bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\} \subseteq + \img{F}{\bigcap\mathscr{A}}.$$ + Let $v \in \bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}$. + Then $\forall A \in \mathscr{A}, v \in \img{F}{A}$. + By definition of the \nameref{ref:image} of a set, + $\forall A \in \mathscr{A}, \exists u \in A, uFv$. + Since $F$ is single-rooted, it follows that + $\exists u, \forall A \in \mathscr{A}, u \in A \land uFv$. + Equivalently, $\exists u \in \bigcap{A}, uFv$. + Thus $v \in \img{F}{\bigcap{A}}$. + +\end{proof} + +\subsection{\partial{Theorem 3K(c)}}% +\label{sub:theorem-3k-c} + +\begin{theorem}[3K(c)] + + The following hold for any sets. ($F$ need not be a function.) + The image of a difference includes the difference of the images: + \begin{equation} + \label{sub:theorem-3k-c-eq1} + \img{F}{A} - \img{F}{B} \subseteq \img{F}{A - B}. + \end{equation} + Equality holds if $F$ is single-rooted. + +\end{theorem} + +\begin{proof} + + We prove that (i) \eqref{sub:theorem-3k-c-eq1} holds and (ii) equality holds + if $F$ is single-rooted. + + \paragraph{(i)}% + \label{par:theorem-3k-c-i} + + Let $v \in \img{F}{A} - \img{F}{B}$. + By definition of the difference of two sets, + $v \in \img{F}{A}$ and $v \not\in \img{F}{B}$. + By definition of the \nameref{ref:image} of a set, there exists a set + $u \in A$ such that $\left< u, v \right> \in F$. + Likewise, $\forall w \in B, \left< w, v \right> \not\in F$. + Thus $u \not\in B$, since otherwise we get an immediate contradiction. + Therefore $u \in A - B$ meaning $v \in \img{F}{A - B}$. + + \paragraph{(ii)}% + + Suppose $F$ is single-rooted. + By \nameref{par:theorem-3k-c-i}, + $$\img{F}{A} - \img{F}{B} \subseteq \img{F}{A - B}.$$ + All that remains is showing + $$\img{F}{A - B} \subseteq \img{F}{A} - \img{F}{B}.$$ + Let $v \in \img{F}{A - B}$. + By definition of the \nameref{ref:image} of a set, there exists a set + $u \in A - B$ such that $uFv$. + Then $u \in A$ and $u \not\in B$. + The former membership relation implies $v \in \img{F}{A}$. + The latter implies $v \not\in \img{F}{B}$ since $F$ being single-rooted + would otherwise invoke an immediate contradiction. + Thus $v \in \img{F}{A} - \img{F}{B}$. + +\end{proof} + +\subsection{\partial{Corollary 3L}}% \label{sub:corollary-3l} \begin{theorem}[3L] For any function $G$ and sets $A$, $B$, and $\mathscr{A}$: - \begin{align*} - G^{-1}\left[\bigcup\mathscr{A}\right] - & = \bigcup\;\{G^{-1}[A] \mid A \in \mathscr{A}\}, \\ + \begin{align} + G^{-1}\left\llbracket\bigcup\mathscr{A}\right\rrbracket + & = \bigcup\;\{G^{-1}[A] \mid A \in \mathscr{A}\}, + \label{sub:corollary-3l-eq1} \\ G^{-1}\left[\bigcap\mathscr{A}\right] & = \bigcap\;\{G^{-1}[A] \mid A \in \mathscr{A}\} - \text{ for } \mathscr{A} \neq \emptyset, \\ - G^{-1}[A - B] - & = G^{-1}[A] - G^{-1}[B]. - \end{align*} + \text{ for } \mathscr{A} \neq \emptyset, + \label{sub:corollary-3l-eq2} \\ + G^{-1}[A - B] & = G^{-1}[A] - G^{-1}[B]. + \label{sub:corollary-3l-eq3} + \end{align} \end{theorem} \begin{proof} - TODO + \nameref{sub:theorem-3k-a} implies \eqref{sub:corollary-3l-eq1}. + Because the inverse of a function is always single-rooted, + \nameref{sub:theorem-3k-b} implies \eqref{sub:corollary-3l-eq2}. + Likewise \nameref{sub:theorem-3k-c} implies \eqref{sub:corollary-3l-eq3}. \end{proof} diff --git a/preamble.tex b/preamble.tex index 46e8ea9..27ca766 100644 --- a/preamble.tex +++ b/preamble.tex @@ -8,6 +8,7 @@ \usepackage{fontawesome5} \usepackage{mathrsfs} \usepackage{soul} +\usepackage{stmaryrd} \usepackage[usenames,dvipsnames]{xcolor} % `hyperref` comes after `xr-hyper`. \usepackage{xr-hyper}