From 3db21d7a1a04cff7575f9b362ad180645fcda902 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Sun, 25 Jun 2023 09:57:03 -0600 Subject: [PATCH] Enderton. Finish remaining theorems 3 set. --- Bookshelf/Enderton/Set.tex | 121 ++++++++++++++++++++++++++++++++++--- 1 file changed, 113 insertions(+), 8 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index b1d81a1..95923b8 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -3221,25 +3221,63 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \end{proof} -\subsection{\unverified{Theorem 3H}}% +\subsection{\partial{Theorem 3H}}% \label{sub:theorem-3h} \begin{theorem}[3H] Assume that $F$ and $G$ are functions. Then $F \circ G$ is a function, its domain is - $$\{x \in \dom{G} \mid G(x) \in \dom{F}\},$$ and for $x$ in its domain, - $(F \circ G)(x) = F(G(x))$. + \begin{equation} + \label{sub:theorem-3h-eq1} + \{x \in \dom{G} \mid G(x) \in \dom{F}\}, + \end{equation} + and for $x$ in its domain, $(F \circ G)(x) = F(G(x))$. \end{theorem} \begin{proof} - TODO + Let $F$ and $G$ be \nameref{ref:function}s. + By definition of the \nameref{ref:composition} of $F$ and $G$, + \begin{equation} + \label{sub:theorem-3h-eq2} + F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}. + \end{equation} + By construction, $F \circ G$ is a relation. + By the definition of the \nameref{ref:domain} of a relation, + $x \in \dom{(F \circ G)}$ if and only if there exists some $y$ such that + $\left< x, y \right> \in F \circ G$. + We prove that (i) $F \circ G$ is a function with domain satisfying + \eqref{sub:theorem-3h-eq1}, and (ii) $(F \circ G)(x) = F(G(x))$. + + \paragraph{(i)}% + \label{par:theorem-3h-i} + + By \eqref{sub:theorem-3h-eq2}, there exists some $t$ such that + $\left< x, t \right> \in G$ and $\left< t, y \right> \in F$. + Since $G$ is single-valued, $t$ is uniquely determined by $x$. + Since $F$ is single-valued, $y$ is uniquely determined by $t$. + Therefore, by transitivity, $y$ is uniquely determined by $x$. + Thus $F \circ G$ is single-valued, i.e. $F \circ G$ is a function. + + Furthermore, by definition of function application, $t = G(x)$. + Thus + $$\left< x, G(x) \right> \in G \quad\text{and}\quad + \left< G(x), y \right> \in F.$$ + This immediately implies \eqref{sub:theorem-3h-eq1} holds true. + + \paragraph{(ii)}% + + Let $x \in \dom{(F \circ G)}$. + By definition, $\left< x, (F \circ G)(x) \right> \in F \circ G$. + Then \eqref{sub:theorem-3h-eq2} implies $(F \circ G)(x)$ satisfies + $\left< G(x), (F \circ G)(x) \right> \in F$. + This is equivalent to saying $F(G(x)) = (F \circ G)(x)$ as expected. \end{proof} -\subsection{\unverified{Theorem 3I}}% +\subsection{\partial{Theorem 3I}}% \label{sub:theorem-3i} \begin{theorem}[3I] @@ -3250,11 +3288,21 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \begin{proof} - TODO + By definition of the \nameref{ref:composition} of $F$ and $G$, + $$F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}.$$ + By definition of the \nameref{ref:inverse} of a function, + \begin{align*} + (F \circ G)^{-1} + & = \{\left< u, v \right> \mid \exists t (vGt \land tFu)\} \\ + & = \{\left< u, v \right> \mid \exists t (tFu \land vGt)\} \\ + & = \{\left< u, v \right> \mid + \exists t \left[ u(F^{-1})t \land t(G^{-1})v \right]\} \\ + & = G^{-1} \circ F^{-1}. + \end{align*} \end{proof} -\subsection{\unverified{Theorem 3J}}% +\subsection{\partial{Theorem 3J}}% \label{sub:theorem-3j} \begin{theorem}[3J] @@ -3273,7 +3321,64 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \begin{proof} - TODO + Let $F$ be a \nameref{ref:function} from nonempty set $A$ to set $B$. + + \paragraph{(a)}% + + We prove there exists a function $G \colon B \rightarrow A$ such that + $G \circ F = I_A$ if and only if $F$ is 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 $y \in \ran{F}$. + 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. + + \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 $F^{-1} \circ F = I_A$. + + \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 $B$. + + \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. + + \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 $F \circ F^{-1} = I_B$. \end{proof}