diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 52c34e0..ef6d7f1 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -24,6 +24,12 @@ \chapter{Reference}% \label{chap:reference} +\section{\partial{Axiom of Choice, First Form}}% +\label{ref:axiom-of-choice-1} + +For any relation $R$ there is a function $H \subseteq R$ with + $\dom{H} = \dom{R}$. + \section{\defined{Composition}}% \label{ref:composition} @@ -3333,7 +3339,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. one-to-one. \item There exists a function $H \colon B \rightarrow A$ (a "right inverse") such that $F \circ H$ is the identity function $I_B$ on $B$ iff $F$ maps - $A$ onto $B$. + $A$ \textit{onto} $B$. \end{enumerate} \end{theorem} @@ -3349,55 +3355,117 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \subparagraph{($\Rightarrow$)}% - Suppose $G \colon B \rightarrow A$ is a function satisfying - $G \circ F = I_A$. - To prove function $F$ is one-to-one, it is sufficient to prove $F$ is - single-rooted. + Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$. + All that remains is to prove $F$ is single-valued. Let $y \in \ran{F}$. - Then there there exists some $x_1 \in \dom{F}$ such that $F(x_1) = y$. - Suppose there exists another $x_2 \in \dom{F}$ such that $F(x_2) = y$. - Then $G(F(x_1)) = G(y) = G(F(x_2))$. - But $G \circ F = I_A$ so $x_1 = x_2$. - Thus, for all $y \in \ran{F}$, there exists only one $x$ such that - $\left< x, y \right> \in F$. - That is, $F$ is single-rooted as expected. + By definition of the \nameref{ref:range} of a function, there exists some + $x$ such that $\left< x, y \right> \in F$. + Suppose $x_1, x_2 \in \dom{F}$ such that + $\left< x_1, y \right>, \left< x_2, y \right> \in F$. + Then $F(x_1) = F(x_2) = y$. + Then $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$. + Thus $x_1 = x_2$. \subparagraph{($\Leftarrow$)}% - Suppose $F$ is one-to-one. - Then by \nameref{sub:lemma-1}, $F^{-1} \colon B \rightarrow A$ is also - one-to-one. - We show that $F^{-1} \circ F = I_A$. - Let $x \in \dom{F}$. - By \nameref{sub:theorem-3h}, $(F^{-1} \circ F)(x) = F^{-1}(F(x))$. - By \nameref{sub:theorem-3g}, $F^{-1}(F(x)) = x$. - Thus $F^{-1} \circ F = I_A$. + Let $F$ be one-to-one. + Since $A$ is nonempty, there exists some $a \in A$. + Let $G \colon B \rightarrow A$ be given by + $$G(y) = \begin{cases} + F^{-1}(y) & \text{if } y \in \ran{F} \\ + a & \text{otherwise}. + \end{cases}$$ + $G$ is a function by virtue of \nameref{sub:lemma-1} and choice of mapping + for all values $y \not\in \ran{F}$. + Furthermore, for all $x \in A$, $F(x) \in \ran{F}$. + Thus $(G \circ F)(x) = G(F(x)) = F^{-1}(F(x)) = x$ by + \nameref{sub:theorem-3g}. \paragraph{(b)}% We prove there exists a function $H \colon B \rightarrow A$ such that - $F \circ H = I_B$ if and only if $F$ maps $A$ onto $B$. + $F \circ H = I_A$ if and only if $F$ maps $A$ onto $B$. \subparagraph{($\Rightarrow$)}% - Suppose $H \colon B \rightarrow A$ is a function satisfying - $F \circ H = I_B$. - To prove $F$ maps $A$ onto $B$, we must prove $\ran{F} = B$. - But for all $y \in B$, $F(H(y)) = y$. - Thus $y \in \ran{F}$ meaning $B \subseteq \ran{F}$. - Since $F$ maps $A$ into $B$ by hypothesis, $\ran{F} \subseteq B$. - Thus $\ran{F} = B$ as expected. + Suppose $H \colon B \rightarrow A$ such that $F \circ H = I_A$. + All that remains is to prove $\ran{F} = B$. + Note that $\ran{F} \subseteq B$ by hypothesis. + Let $y \in B$. + But $F(H(y)) = y$ meaning $y \in \ran{F}$. + Thus $B \subseteq \ran{F}$. + Since $\ran{F} \subseteq B$ and $B \subseteq \ran{F}$, $\ran{F} = B$. \subparagraph{($\Leftarrow$)}% - Suppose $F$ maps $A$ onto $B$. - That is, $\ran{F} = B$. - We show that $F^{-1} \colon B \rightarrow A$ satisfies - $F \circ F^{-1} = I_B$. - Let $y \in \ran{F}$. - By \nameref{sub:theorem-3h}, $(F \circ F^{-1})(y) = F(F^{-1}(y))$. - By \nameref{sub:theorem-3g}, $F(F^{-1}(y)) = y$. - Thus $F \circ F^{-1} = I_B$. + Suppose $F$ maps $A$ \textit{onto} $B$. + By definition of maps onto, $\ran{F} = B$. + Then for all $y \in B$, there exists some $x \in A$ such that + $\left< x, y \right> \in F$. + Notice though that $F^{-1}[\{y\}]$ may not be a singleton set. + Then there is no obvious way to \textit{choose} an element from each + preimage to form a function. + By the \nameref{ref:axiom-of-choice-1}, there exists a function + $H \subseteq F^{-1}$ such that $\dom{H} = \dom{F^{-1}} = B$. + For all $y \in B$, $\left< y, H(y) \right> \in H \subseteq F^{-1}$ + meaning $\left< H(y), y \right> \in F$. + Thus $F(H(y)) = y$ as expected. + +\end{proof} + +\subsection{\unverified{Theorem 3K}}% +\label{sub:theorem-3k} + +\begin{theorem}[3K] + + 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} + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{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}\}, \\ + 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*} + +\end{theorem} + +\begin{proof} + + TODO \end{proof}