From 738435e29973b9efa674114ac082f0abb7af9525 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Fri, 16 Jun 2023 08:16:33 -0600 Subject: [PATCH] Add prompts for upcoming theorems/exercises. --- Bookshelf/Enderton/Set.tex | 261 ++++++++++++++++++++++++-- Bookshelf/Enderton/Set/Chapter_3.lean | 18 ++ 2 files changed, 266 insertions(+), 13 deletions(-) diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index b352a0f..d0e7400 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -6,6 +6,10 @@ \input{../../preamble} \makeleancommands{../..} +\newcommand{\dom}[1]{\textop{dom}{#1}} +\newcommand{\fld}[1]{\textop{fld}{#1}} +\newcommand{\ran}[1]{\textop{ran}{#1}} + \begin{document} \header{Elements of Set Theory}{Herbert B. Enderton} @@ -45,6 +49,32 @@ If two sets have exactly the same members, then they are equal: \end{axiom} +\section{\defined{Function}}% +\label{ref:function} + +A \textbf{function} is a relation $F$ such that for each $x$ in $\dom{F}$ there + is only one $y$ such that $xFy$. +We say that $F$ is a function \textbf{from $A$ into $B$} or that $F$ + \textbf{maps $A$ into $B$} (written $F \colon A \rightarrow B$) iff $F$ is a + function, $\dom{F} = A$, and $\ran{F} \subseteq B$. +If $\ran{F} = B$, then $F$ is a function from \textbf{$A$ onto $B$}. + +A function $F$ is \textbf{one-to-one} iff for each $y \in \ran{F}$ there is only + one $x$ such that $xFy$. +One-to-one functions are sometimes called \textbf{injections}. + +\begin{definition} + + \statementpadding + + \lean*{Mathlib/Init/Function}{Function.Injective} + + \lean*{Mathlib/Init/Function}{Function.Surjective} + + \lean*{Mathlib/Init/Function}{Function.Bijective} + +\end{definition} + \section{\defined{Ordered Pair}}% \label{ref:ordered-pair} @@ -114,6 +144,30 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$: \end{axiom} +\section{\defined{Relation}}% +\label{ref:relation} + +A \textbf{relation} is a set of \nameref{ref:ordered-pair}s. +Given relation $R$, the \textbf{domain} of $R$ ($\dom{R}$), the \textbf{range} + of $R$ ($\ran{R}$), and the \textbf{field} of $R$ ($\fld{R}$) is given by + \begin{align*} + x \in \dom{R} & \iff \exists y \left< x, y \right> \in R, \\ + x \in \ran{R} & \iff \exists t \left< t, x \right> \in R, \\ + \fld{R} & = \dom{R} \cup \ran{R}. + \end{align*} + +\begin{definition} + + \statementpadding + + \lean*{Mathlib/Data/Rel}{Rel} + + \lean*{Mathlib/Data/Rel}{Rel.dom} + + \lean*{Mathlib/Data/Rel}{Rel.codom} + +\end{definition} + \section{\defined{Subset Axioms}}% \label{ref:subset-axioms} @@ -2656,26 +2710,82 @@ With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$. \section{Exercises 6}% \label{sec:exercises-6} -\subsection{\unverified{Exercise 6.6}}% +\subsection{\verified{Exercise 6.6}}% \label{sub:exercise-6.6} -Show that a set $A$ is a relation iff - $A \subseteq \textop{dom} A \times \textop{ran} A$. +Show that a set $A$ is a relation iff $A \subseteq \dom{A} \times \ran{A}$. \begin{proof} - TODO + \lean{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.exercise\_6\_6} + + Let $A$ be a set. + We prove the forward and reverse direction of the bidirectional. + + \paragraph{($\Rightarrow$)}% + + Suppose $A$ is a \nameref{ref:relation}. + We show for all $a \in A$, $a \in \dom{A} \times \ran{A}$. + Let $a \in A$. + Since $A$ is a relation, $a$ is an ordered pair. + Then there exists some sets $x$ and $y$ such that $a = \left< x, y \right>$. + By definition of the domain and range of $A$, $x \in \dom{A}$ and + $y \in \ran{A}$. + Thus $a = \left< x, y \right> \in \dom{A} \times \ran{A}$ as well. + This proves $A \subseteq \dom{A} \times \ran{A}$. + + \paragraph{($\Leftarrow$)}% + + Suppose $A \subseteq \dom{A} \times \ran{A}$. + Then for all $a \in A$, $a \in \dom{A} \times \ran{A}$. + Therefore $a$ is an ordered pair. + Since this holds for all $a \in A$, it follows $A$ is a relation. \end{proof} -\subsection{\unverified{Exercise 6.7}}% +\subsection{\partial{Exercise 6.7}}% \label{sub:exercise-6.7} -Show that if $R$ is a relation, then $\textop{fld} R = \bigcup\bigcup R$. +Show that if $R$ is a relation, then $\fld{R} = \bigcup\bigcup R$. \begin{proof} - TODO + Let $R$ be a \nameref{ref:relation}. + We show that (i) $\fld{R} \subseteq \bigcup\bigcup R$ and (ii) that + $\bigcup\bigcup R \subseteq \fld{R}$. + + \paragraph{(i)}% + \label{par:exercise-6.7-i} + + Let $x \in \fld{R} = \dom{R} \cup \ran{R}$. + That is, $x \in \dom{R}$ or $x \in \ran{R}$. + + If $x \in \dom{R}$, then there exists some $y$ such that + $\left< x, y \right> = \{\{x\}, \{x, y\}\} \in R$. + Then $\{x\} \in \bigcup R$ and $x \in \bigcup\bigcup R$. + + On the other hand, if $x \in \ran{R}$, then there exists some $t$ such that + $\left< t, x \right> = \{\{t\}, \{t, x\}\} \in R$. + Then $\{t, x\} \in \bigcup R$ and $x \in \bigcup\bigcup R$. + + \paragraph{(ii)}% + \label{par:exercise-6.7-ii} + + Let $t \in \bigcup\bigcup R$. + Then there exists some member $T \in \bigcup R$ such that $t \in T$. + Likewise there exists some member $T' \in R$ such that $T \in T'$. + By definition of a relation, + $T' = \left< x, y \right> = \{\{x\}, \{x, y\}\}$ for some sets $x$ and + $y$. + Thus $t = x$ or $t = y$. + By \nameref{sub:exercise-6.6}, $t \in \dom{R}$ or $t \in \ran{R}$. + In other words, $t \in \fld{R}$. + + \paragraph{Conclusion}% + + Since \nameref{par:exercise-6.7-i} and \nameref{par:exercise-6.7-ii} hold, + $\fld{R} = \bigcup\bigcup{R}$. \end{proof} @@ -2683,12 +2793,14 @@ Show that if $R$ is a relation, then $\textop{fld} R = \bigcup\bigcup R$. \label{sub:exercise-6.8} Show that for any set $\mathscr{A}$: - \begin{align*} - \textop{dom} \bigcup{\mathscr{A}} - & = \bigcup\; \{ \textop{dom} R \mid R \in \mathscr{A} \}, - \textop{ran} \bigcup{\mathscr{A}} - & = \bigcup\; \{ \textop{ran} R \mid R \in \mathscr{A} \}. - \end{align*} + \begin{align} + \dom{\bigcup{\mathscr{A}}} + & = \bigcup\; \{ \dom{R} \mid R \in \mathscr{A} \}, + & \label{sub:exercise-6.8-eq1} \\ + \ran{\bigcup{\mathscr{A}}} + & = \bigcup\; \{ \ran{R} \mid R \in \mathscr{A} \}. + & \label{sub:exercise-6.8-eq2} + \end{align} \begin{proof} @@ -2708,4 +2820,127 @@ Discuss the result of replacing the union operation by the intersection \end{proof} +\section{Exercises 7}% +\label{sec:exercises-7} + +\subsection{\unverified{Exercise 7.10}}% +\label{sub:exercise-7.10} + +Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive + integer $m$ less than $4$. + +\begin{proof} + + TODO + +\end{proof} + +\section{Functions}% +\label{sec:functions} + +\subsection{\unverified{Theorem 3E}}% +\label{sub:theorem-3e} + +\begin{theorem}[3E] + + For a set $F$, $\dom{F^{-1}} = \ran{F}$ and $\ran{F^{-1}} = \dom{F}$. + For a relation $F$, $(F^{-1})^{-1} = F$. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Theorem 3F}}% +\label{sub:theorem-3f} + +\begin{theorem}[3F] + + For a set $F$, $F^{-1}$ is a function iff $F$ is single-rooted. + A relation $F$ is a function iff $F^{-1}$ is single-rooted. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Theorem 3G}}% +\label{sub:theorem-3g} + +\begin{theorem}[3G] + + Assume that $F$ is a one-to-one function. + If $x \in \dom{F}$, then $F^{-1}(F(x)) = x$. + If $y \in \ran{F}$, then $F(F^{-1}(y)) = y$. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{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, + $(F \circ G)(x) = F(G(x))$. + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Theorem 3I}}% +\label{sub:theorem-3i} + +\begin{theorem}[3I] + + For any sets $F$ and $G$, $$(F \circ G)^{-1} = G^{-1} \circ F^{-1}.$$ + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + +\subsection{\unverified{Theorem 3J}}% +\label{sub:theorem-3j} + +\begin{theorem}[3J] + + Assume that $F \colon A \rightarrow B$, and that $A$ is nonempty. + \begin{enumerate}[(a)] + \item There exists a function $G \colon B \rightarrow A$ (a "left inverse") + such that $G \circ F$ is the identity function $I_A$ on $A$ iff $F$ is + 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 $B$. + \end{enumerate} + +\end{theorem} + +\begin{proof} + + TODO + +\end{proof} + \end{document} diff --git a/Bookshelf/Enderton/Set/Chapter_3.lean b/Bookshelf/Enderton/Set/Chapter_3.lean index e41790f..c079bc7 100644 --- a/Bookshelf/Enderton/Set/Chapter_3.lean +++ b/Bookshelf/Enderton/Set/Chapter_3.lean @@ -328,4 +328,22 @@ theorem theorem_3d {A : Set (Set (Set (α ⊕ α)))} (h : OrderedPair x y ∈ A) have : ∀ t, t ∈ {Sum.inl x, Sum.inr y} → t ∈ ⋃₀ (⋃₀ A) := hq' exact ⟨this (Sum.inl x) (by simp), this (Sum.inr y) (by simp)⟩ +/-- ### Exercise 6.6 + +Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`. +-/ +theorem exercise_6_6 {A : Set (α × β)} + : A ⊆ Set.prod (Prod.fst '' A) (Prod.snd '' A) := by + show ∀ t, t ∈ A → t ∈ Set.prod (Prod.fst '' A) (Prod.snd '' A) + intro (a, b) ht + unfold Set.prod + simp only [ + Set.mem_image, + Prod.exists, + exists_and_right, + exists_eq_right, + Set.mem_setOf_eq + ] + exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩ + end Enderton.Set.Chapter_3 \ No newline at end of file