\documentclass{report} \usepackage{graphicx} \graphicspath{{./Set/images/}} \input{../../preamble} \makecode{../..} \externaldocument[S:]{../../Common/Real/Sequence} \newcommand{\ineq}{\,\mathop{\underline{\in}}\,} \begin{document} \header{Elements of Set Theory}{Herbert B. Enderton} \tableofcontents \begingroup \renewcommand\thechapter{R} \setcounter{chapter}{0} \addtocounter{chapter}{-1} \chapter{Reference}% \hyperlabel{chap:reference} \section{\defined{Addition}}% \hyperlabel{ref:addition} For each $m \in \omega$, there exists (by the \nameref{sub:recursion-theorem-natural-numbers}) a unique \nameref{ref:function} $A_m \colon \omega \rightarrow \omega$ for which \begin{align*} A_m(0) & = m, \\ A_m(n^+) & = A_m(n)^+ & \text{for } n \text{ in } \omega. \end{align*} \textbf{Addition} ($+$) is the \nameref{ref:binary-operation} on $\omega$ such that for any $m$ and $n$ in $\omega$, $$m + n = A_m(n).$$ \lean{Init/Prelude} {Add.add} \section{\defined{Axiom of Choice, First Form}}% \hyperlabel{ref:axiom-of-choice-1} For any relation $R$ there is a function $H \subseteq R$ with $\dom{H} = \dom{R}$. \lean*{Init/Prelude} {Classical.choice} \section{\defined{Axiom of Choice, Second Form}}% \hyperlabel{ref:axiom-of-choice-2} For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$ for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$ \lean{Init/Prelude} {Classical.choice} \section{\defined{Binary Operation}}% \hyperlabel{ref:binary-operation} A \textbf{binary operation} on a set $A$ is a \nameref{ref:function} from $A \times A$ into $A$. \section{\defined{Cardinal Arithmetic}}% \hyperlabel{sec:cardinal-arithmetic} Let $\kappa$ and $\lambda$ be any cardinal numbers. \begin{enumerate}[(a)] \item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any disjoint sets of cardinality $\kappa$ and $\lambda$, respectively. \item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are any sets of cardinality $\kappa$ and $\lambda$, respectively. \item $\kappa^\lambda = \card{(^L{K})}$, where $K$ and $L$ are any sets of cardinality $\kappa$ and $\lambda$, respectively. \end{enumerate} \lean{Mathlib/SetTheory/Cardinal/Basic} {Cardinal.add\_def} \lean{Mathlib/SetTheory/Cardinal/Basic} {Cardinal.mul\_def} \lean{Mathlib/SetTheory/Cardinal/Basic} {Cardinal.power\_def} \section{\defined{Cardinal Number}}% \hyperlabel{ref:cardinal-number} For any set $C$, the \textbf{cardinal number} of set $C$ is denoted as $\card{C}$. Furthermore, \begin{enumerate}[(a)] \item For any sets $A$ and $B$, $$\card{A} = \card{B} \quad\text{iff}\quad A \equin B.$$ \item For a finite set $A$, $\card{A}$ is the \nameref{ref:natural-number} $n$ for which $A \equin n$. \end{enumerate} \lean{Mathlib/Data/Finset/Card} {Finset.card} \lean{Mathlib/SetTheory/Cardinal/Basic} {Cardinal} \section{\defined{Cartesian Product}}% \hyperlabel{ref:cartesian-product} Let $I$ be a set and let $H$ be a \nameref{ref:function} whose domain includes $I$. Then for each $i$ in $I$ we have the set $H(i)$. We define the \textbf{cartesian product} of the $H(i)$'s as $$\bigtimes_{i \in I} H(i) = \{f \mid f \text{ is a function with domain } I \text{ and } (\forall i \in I) f(i) \in H(i)\}.$$ \lean{Mathlib/Data/Set/Prod}{Set.prod} \section{\defined{Compatible}}% \hyperlabel{ref:compatible} A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and only if for all $x$ and $y$ in $A$, $$xRy \Rightarrow F(x)RF(y).$$ \lean{Init/Core} {Quotient.lift} \section{\defined{Composition}}% \hyperlabel{ref:composition} The \textbf{composition} of sets $F$ and $G$ is $$F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}.$$ \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.comp} \lean{Mathlib/Data/Rel} {Rel.comp} \section{\defined{Connected}}% \hyperlabel{ref:connected} A binary relation $R$ on $A$ is \textbf{connected} if for distinct $x, y \in A$, either $xRy$ or $yRx$. \lean*{Common/Algebra/Classes}{IsConnected} \section{\defined{Domain}}% \hyperlabel{ref:domain} The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by $$x \in \dom{R} \iff \exists y \tuple{x, y} \in R.$$ \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.dom} \lean{Mathlib/Data/Rel} {Rel.dom} \section{\defined{Empty Set Axiom}}% \hyperlabel{ref:empty-set-axiom} There is a set having no members: $$\exists B, \forall x, x \not\in B.$$ \lean{Mathlib/Init/Set}{Set.emptyCollection} \section{\defined{Equinumerous}}% \hyperlabel{ref:equinumerous} A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \equin B$) if and only if there is a one-to-one \nameref{ref:function} from $A$ onto $B$. In other words, there exists a one-to-one correspondence between $A$ and $B$. \lean*{Mathlib/Init/Function} {Function.Bijective} \lean{Mathlib/Data/Set/Function} {Set.BijOn} \lean{Mathlib/Logic/Equiv/Defs} {Equiv} \section{\defined{Equivalence Class}}% \hyperlabel{ref:equivalence-class} The set $[x]_R$ is defined by $$[x]_R = \{t \mid xRt\}.$$ If $R$ is an \nameref{ref:equivalence-relation} and $x \in \fld{R}$, then $[x]_R$ is called the \textbf{equivalence class} of $x$ (\textbf{modulo $R$}). If the relation $R$ is fixed by the context, we may write just $[x]$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.modEquiv} \section{\defined{Equivalence Relation}}% \hyperlabel{ref:equivalence-relation} Relation $R$ is an \textbf{equivalence relation} on set $A$ if and only if $R$ is a binary \nameref{ref:relation} on $A$ that is \nameref{ref:reflexive} on $A$, \nameref{ref:symmetric}, and \nameref{ref:transitive}. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.isEquivalence} \section{\defined{Exponentiation}}% \hyperlabel{ref:exponentiation} For each $m \in \omega$, there exists (by the \nameref{sub:recursion-theorem-natural-numbers}) a unique \nameref{ref:function} $E_m \colon \omega \rightarrow \omega$ for which \begin{align*} E_m(0) & = 1, \\ E_m(n^+) & = E_m(n) \cdot m & \text{for } n \text{ in } \omega. \end{align*} \textbf{Exponentiation} is the \nameref{ref:binary-operation} on $\omega$ such that for any $m$ and $n$ in $\omega$, $$m^n = E_m(n).$$ \lean{Init/Prelude} {Pow.pow} \section{\defined{Extensionality Axiom}}% \hyperlabel{ref:extensionality-axiom} If two sets have exactly the same members, then they are equal: $$\forall A, \forall B, \left[\forall x, (x \in A \iff x \in B) \Rightarrow A = B\right].$$ \lean{Mathlib/Init/Set} {Set.ext} \section{\defined{Field}}% \hyperlabel{ref:field} Given \nameref{ref:relation} $R$, the \textbf{field} of $R$, denoted $\fld{R}$, is given by $$\fld{R} = \dom{R} \cup \ran{R}.$$ \lean{Bookshelf/Enderton/Set/Relation} {Set.Relation.fld} \section{\defined{Finite Set}}% \hyperlabel{ref:finite-set} A set is \textbf{finite} if and only if it is \nameref{ref:equinumerous} to a \nameref{ref:natural-number}. \code*{Common/Set/Finite} {Set.finite\_iff\_equinumerous\_nat} \lean{Mathlib/Data/Set/Finite} {Set.Finite} \section{\defined{Function}}% \hyperlabel{ref:function} A \textbf{function} is a relation $F$ such that for each $x$ in $\dom{F}$ there is only one $y$ such that $xFy$. In other words, $F$ is \textbf{single-valued}. 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}. A one-to-one function from $A$ onto $B$ is a \textbf{one-to-one correspondence} between $A$ and $B$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.isSingleValued} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.isSingleRooted} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.isOneToOne} \lean{Mathlib/Data/Set/Function} {Set.MapsTo} \lean{Mathlib/Data/Set/Function} {Set.InjOn} \lean{Mathlib/Data/Set/Function} {Set.SurjOn} \lean{Mathlib/Data/Set/Function} {Set.BijOn} \section{\defined{Image}}% \hyperlabel{ref:image} Let $A$ and $F$ be arbitrary sets. The \textbf{image of $A$ under $F$} is the set \begin{align*} \img{F}{A} & = \ran{(F \restriction A)} \\ & = \{v \mid (\exists u \in A) uFv\}. \end{align*} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.image} \lean{Mathlib/Data/Rel} {Rel.image} \section{\defined{Inductive Set}}% \hyperlabel{ref:inductive-set} A set $A$ is said to be \textit{inductive} if and only if $\emptyset \in A$ and it is "closed under \nameref{ref:successor}", i.e. $$(\forall a \in A) a^+ \in A.$$ \lean{Prelude} {Nat} \lean{Mathlib/Init/Set} {Set.univ} \begin{note} Induction is baked into Lean's type system. In particular, the $\emptyset$ and "closed under successor" properties are analagous to base and recursive constructors of an inductive data type respectively. \end{note} \section{\defined{Infinite Set}}% \hyperlabel{ref:infinite-set} A set is \textbf{infinite} if and only if it is not a \nameref{ref:finite-set}. \lean*{Mathlib/Data/Set/Finite} {Set.Infinite} \section{\defined{Infinity Axiom}}% \hyperlabel{ref:infinity-axiom} There exists an \nameref{ref:inductive-set}: $$(\exists A)\left[ \emptyset \in A \land (\forall a \in A) a^+ \in A\right].$$ \lean{Prelude} {Nat} \lean{Mathlib/Init/Set} {Set.univ} \section{\defined{Inverse}}% \hyperlabel{ref:inverse} The \textbf{inverse} of a set $F$ is the set $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$ \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.inv} \lean{Mathlib/Data/Rel} {Rel.inv} \section{\defined{Irreflexive}}% \hyperlabel{ref:irreflexive} A binary relation $R$ on set $A$ is \textbf{irreflexive} if there is no $x \in A$ for which $xRx$. \lean*{Mathlib/Init/Algebra/Classes} {IsIrrefl} \section{\defined{Linear Ordering}} \hyperlabel{ref:linear-ordering} Let $A$ be any set. A \textbf{linear ordering} on $A$ (also called a \textbf{total ordering} on $A$) is a binary relation $R$ on $A$ (i.e., $R \subseteq A \times A$) meeting the following two conditions: \begin{enumerate}[(a)] \item $R$ is \nameref{ref:transitive}. \item $R$ is \nameref{ref:trichotomous}. \end{enumerate} \lean{Mathlib/Init/Algebra/Classes} {IsStrictTotalOrder} \begin{note} This definition does not agree with how Lean defines a linear order. \vspace{6pt} Trichotomy is equivalent to asymmetry and connectivity and asymmetry is equivalent to antisymmetry and irreflexivity. Thus a linear order, as defined by Enderton, is a binary relation with the following four properties: \vspace{6pt} \begin{enumerate}[(i)] \item Irreflexivity \item Antisymmetry \item Connectivity (i.e. totality) \item Transitivity \end{enumerate} \end{note} \section{\defined{Multiplication}}% \hyperlabel{ref:multiplication} For each $m \in \omega$, there exists (by the \nameref{sub:recursion-theorem-natural-numbers}) a unique \nameref{ref:function} $M_m \colon \omega \rightarrow \omega$ for which \begin{align*} M_m(0) & = 0, \\ M_m(n^+) & = M_m(n) + m. \end{align*} \textbf{Multiplication} ($\cdot$) is the \nameref{ref:binary-operation} on $\omega$ such that for any $m$ and $n$ in $\omega$, $$m \cdot n = M_m(n).$$ \lean{Init/Prelude} {Mul.mul} \section{\defined{Natural Map}}% \hyperlabel{ref:natural-map} Let $R$ be an \nameref{ref:equivalence-relation} on $A$. Then the \textbf{natural map} (or \textbf{canonical map}) $\phi \colon A \rightarrow A / R$ is defined as $$\phi(x) = [x]_R$$ for $x \in A$. \lean*{Init/Core} {Quotient.lift} \section{\defined{Natural Number}}% \hyperlabel{ref:natural-number} A \textbf{natural number} is a set that belongs to every inductive set. The set of all natural numbers exists by virtue of \nameref{sub:theorem-4a}. This set is denoted as $\omega$. \lean*{Prelude} {Nat} \section{\defined{Ordered Pair}}% \hyperlabel{ref:ordered-pair} For any sets $u$ and $v$, the \textbf{ordered pair} $\tuple{u, v}$ is the set $\{\{u\}, \{u, v\}\}$. \code*{Bookshelf/Enderton/Set/OrderedPair} {OrderedPair} \lean{Prelude} {Prod} \section{\defined{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}% \hyperlabel{ref:ordering-natural-numbers} For \nameref{ref:natural-number}s $m$ and $n$, define $m$ to be \textbf{less than} $n$ if and only if $m \in n$. That is, $$m < n \iff m \in n.$$ Likewise, define $m$ to be \textbf{less than or equal to} $n$ if and only if $m \in n \lor m = n$. That is, \begin{align*} m \leq n & \iff m \ineq n \\ & \iff m < n \lor m = n. \end{align*} \lean{Init/Prelude} {Nat.lt} \section{\defined{Pair Set}}% \hyperlabel{ref:pair-set} For any sets $u$ and $v$, the \textbf{pair set $\{u, v\}$} is the set whose only members are $u$ and $v$. \lean*{Mathlib/Init/Set} {Set.insert} \lean{Mathlib/Init/Set} {Set.singleton} \section{\defined{Pairing Axiom}}% \hyperlabel{ref:pairing-axiom} For any sets $u$ and $v$, there is a set having as members just $u$ and $v$: $$\forall u, \forall v, \exists B, \forall x, (x \in B \iff x = u \text{ or } x = v).$$ \lean{Mathlib/Init/Set} {Set.insert} \lean{Mathlib/Init/Set} {Set.singleton} \section{\defined{Partition}}% \hyperlabel{ref:partition} A \textbf{partition} $\Pi$ of a set $A$ is a set of nonempty subsets of $A$ that is disjoint and exhaustive, i.e. \begin{enumerate}[(a)] \item no two different sets in $\Pi$ have any common elements, and \item each element of $A$ is in some set in $\Pi$. \end{enumerate} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.Partition} \lean{Mathlib/Data/Setoid/Partition} {Setoid.IsPartition} \section{\defined{Peano System}}% \hyperlabel{ref:peano-system} A \textbf{Peano system} is a triple $\langle N, S, e \rangle$ consisting of a set $N$, a function $S \colon N \rightarrow N$, and a member $e \in N$ such that the following three conditions are met: \begin{enumerate}[(i)] \item $e \not\in \ran{S}$. \item $S$ is one-to-one. \item Every subset $A$ of $N$ containing $e$ and closed under $S$ is $N$ itself. \end{enumerate} \code{Common/Set/Peano} {Peano.System} \section{\defined{Power Set}}% \hyperlabel{ref:power-set} For any set $a$, the \textbf{power set $\powerset{a}$} is the set whose members are exactly the subsets of $a$. \lean*{Mathlib/Init/Set} {Set.powerset} \section{\defined{Power Set Axiom}}% \hyperlabel{ref:power-set-axiom} For any set $a$, there is a set whose members are exactly the subsets of $a$: $$\forall a, \exists B, \forall x, (x \in B \iff x \subseteq a).$$ \lean{Mathlib/Init/Set} {Set.powerset} \section{\defined{Proper Subset}}% \hyperlabel{ref:proper-subset} A set $A$ is said to be a \textbf{proper subset} of $B$ ($A \subset B$) if and only if it is a subset of $B$ that is unequal to $B$. $$A \subset B \iff A \subseteq B \land A \neq B.$$ \lean{Std/Classes/SetNotation} {HasSSubset} \section{\defined{Quotient Set}}% \hyperlabel{ref:quotient-set} If $R$ is an \nameref{ref:equivalence-relation} on set $A$, then we can define the \textbf{quotient set} $$A / R = \{[x]_R \mid x \in A\}$$ whose members are the equivalence classes. The expression $A / R$ is read "$A$ modulo $R$. \lean*{Init/Core} {Quotient} \section{\defined{Range}}% \hyperlabel{ref:range} The \textbf{range} of set $R$, denoted $\ran{R}$, is given by $$x \in \ran{R} \iff \exists t \tuple{t, x} \in R.$$ \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.ran} \lean{Mathlib/Data/Rel} {Rel.codom} \section{\defined{Reflexive}}% \hyperlabel{ref:reflexive} A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for all $x \in A$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.isReflexive} \lean{Mathlib/Init/Algebra/Classes} {IsRefl} \section{\defined{Relation}}% \hyperlabel{ref:relation} A \textbf{relation} is a set of \nameref{ref:ordered-pair}s. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation} \lean{Mathlib/Data/Rel}{Rel} \section{\defined{Restriction}}% \hyperlabel{ref:restriction} The \textbf{restriction} of a set $F$ to set $A$ is the set $$F \restriction A = \{\tuple{u, v} \mid uFv \land u \in A\}.$$ \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.restriction} \section{\defined{Successor}}% \hyperlabel{ref:successor} For any set $a$, its \textbf{successor} is defined by $$a^+ = a \cup \{a\}.$$ \lean{Prelude} {Nat.succ} \begin{note} The corresponding Lean definition refers to the `Nat.succ` constructor. This is not represented internally as a union of sets, but serves the same role. \end{note} \section{\defined{Subset Axioms}}% \hyperlabel{ref:subset-axioms} For each formula $\phi$ not containing $B$, the following is an axiom: $$\forall t_1, \cdots \forall t_k, \forall c, \exists B, \forall x, (x \in B \iff x \in c \land \phi).$$ \lean{Mathlib/Init/Set} {Set.Subset} \section{\defined{Symmetric}}% \hyperlabel{ref:symmetric} A binary relation $R$ is \textbf{symmetric} if and only if whenever $xRy$ then $yRx$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.isSymmetric} \section{\defined{Symmetric Difference}}% \hyperlabel{ref:symmetric-difference} The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set $(A - B) \cup (B - A)$. \lean*{Mathlib/Data/Set/Basic} {symmDiff\_def} \section{\defined{Transitive}}% \hyperlabel{ref:transitive} A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and $yRz$, then $xRz$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.isTransitive} \lean{Mathlib/Init/Algebra/Classes} {IsTrans} \section{\defined{Transitive Set}}% \hyperlabel{ref:transitive-set} A set $A$ is said to be \textbf{transitive} if and only if every member of a member of $A$ is a member of $A$ itself. That is, $\bigcup A \subseteq A$. \section{\defined{Trichotomous}}% \hyperlabel{ref:trichotomous} A binary relation $R$ on set $A$ is \textbf{trichotomous} if for any $x, y \in A$, exactly one of the three alternatives $$xRy, \quad x = y, \quad yRx$$ holds. \lean*{Mathlib/Init/Algebra/Classes} {IsTrichotomous} \section{\defined{Union Axiom}}% \hyperlabel{ref:union-axiom} For any set $A$, there exists a set $B$ whose elements are exactly the members of the members of $A$: $$\forall A, \exists B, \forall x \left[ x \in B \iff (\exists b \in A) x \in b \right]$$ \lean{Mathlib/Data/Set/Lattice} {Set.sUnion} \section{\defined{Union Axiom, Preliminary Form}}% \hyperlabel{ref:union-axiom-preliminary-form} For any sets $a$ and $b$, there is a set whose members are those sets belonging either to $a$ or to $b$ (or both): $$\forall a, \forall b, \exists B, \forall x, (x \in B \iff x \in a \text{ or } x \in b).$$ \lean{Mathlib/Init/Set} {Set.union} \endgroup \chapter{Introduction}% \hyperlabel{chap:introduction} \section{Exercises 1}% \hyperlabel{sec:exercises-1} \subsection{\verified{Exercise 1.1}}% \hyperlabel{sub:exercise-1.1} Which of the following become true when "$\in$" is inserted in place of the blank? Which become true when "$\subseteq$" is inserted? \subsubsection{\verified{Exercise 1.1a}}% \hyperlabel{ssub:exercise-1.1a} $$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}.$$ \code{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_1a} \begin{proof} Because the \textit{object} $\{\emptyset\}$ is a member of the right-hand set, the statement is \textbf{true} in the case of "$\in$". Because the \textit{members} of $\{\emptyset\}$ are all members of the right-hand set, the statement is also \textbf{true} in the case of "$\subseteq$". \end{proof} \subsubsection{\verified{Exercise 1.1b}}% \hyperlabel{ssub:exercise-1.11b} $$\{\emptyset\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}.$$ \code{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_1b} \begin{proof} Because the \textit{object} $\{\emptyset\}$ is not a member of the right-hand set, the statement is \textbf{false} in the case of "$\in$". Because the \textit{members} of $\{\emptyset\}$ are all members of the right-hand set, the statement is \textbf{true} in the case of "$\subseteq$". \end{proof} \subsubsection{\verified{Exercise 1.1c}}% \hyperlabel{ssub:exercise-1.1c} $$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\emptyset\}\}.$$ \code{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_1c} \begin{proof} Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the right-hand set, the statement is \textbf{false} in the case of "$\in$". Because the \textit{members} of $\{\{\emptyset\}\}$ are all members of the right-hand set, the statement is \textbf{true} in the case of "$\subseteq$". \end{proof} \subsubsection{\verified{Exercise 1.1d}}% \hyperlabel{ssub:exercise-1.1d} $$\{\{\emptyset\}\} \_\_\_\_ \{\emptyset, \{\{\emptyset\}\}\}.$$ \code{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_1d} \begin{proof} Because the \textit{object} $\{\{\emptyset\}\}$ is a member of the right-hand set, the statement is \textbf{true} in the case of "$\in$". Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the right-hand set, the statement is \textbf{false} in the case of "$\subseteq$". \end{proof} \subsubsection{\verified{Exercise 1.1e}}% \hyperlabel{ssub:exercise-1.1e} $$\{\{\emptyset\}\} \_\_ \{\emptyset, \{\emptyset, \{\emptyset\}\}\}.$$ \code{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_1e} \begin{proof} Because the \textit{object} $\{\{\emptyset\}\}$ is not a member of the right-hand set, the statement is \textbf{false} in the case of "$\in$". Because the \textit{members} of $\{\{\emptyset\}\}$ are not all members of the right-hand set, the statement is \textbf{false} in the case of "$\subseteq$". \end{proof} \subsection{\verified{Exercise 1.2}}% \hyperlabel{sub:exercise-1.2} Show that no two of the three sets $\emptyset$, $\{\emptyset\}$, and $\{\{\emptyset\}\}$ are equal to each other. \code*{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_2} \begin{proof} By the \nameref{ref:extensionality-axiom}, $\emptyset$ is only equal to $\emptyset$. This immediately shows it is not equal to the other two. Now consider object $\emptyset$. This object is a member of $\{\emptyset\}$ but is not a member of $\{\{\emptyset\}\}$. Again, by the \nameref{ref:extensionality-axiom}, these two sets must be different. \end{proof} \subsection{\verified{Exercise 1.3}}% \hyperlabel{sub:exercise-1.3} Show that if $B \subseteq C$, then $\powerset{B} \subseteq \powerset{C}$. \code*{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_3} \begin{proof} Let $x \in \powerset{B}$. By definition of the \nameref{ref:power-set}, $x$ is a subset of $B$. By hypothesis, $B \subseteq C$. Then $x \subseteq C$. Again by definition of the \nameref{ref:power-set}, it follows $x \in \powerset{C}$. \end{proof} \subsection{\verified{Exercise 1.4}}% \hyperlabel{sub:exercise-1.4} Assume that $x$ and $y$ are members of a set $B$. Show that $\{\{x\}, \{x, y\}\} \in \powerset{\powerset{B}}.$ \code*{Bookshelf/Enderton/Set/Chapter\_1} {Enderton.Set.Chapter\_1.exercise\_1\_4} \begin{proof} Let $x$ and $y$ be members of set $B$. Then $\{x\}$ and $\{x, y\}$ are subsets of $B$. By definition of the \nameref{ref:power-set}, $\{x\}$ and $\{x, y\}$ are members of $\powerset{B}$. Then $\{\{x\}, \{x, y\}\}$ is a subset of $\powerset{B}$. By definition of the \nameref{ref:power-set}, $\{\{x\}, \{x, y\}\}$ is a member of $\powerset{\powerset{B}}$. \end{proof} \subsection{\unverified{Exercise 1.5}}% \hyperlabel{sub:exercise-1.5} Define the rank of a set $c$ to be the least $\alpha$ such that $c \subseteq V_\alpha$. Compute the rank of $\{\{\emptyset\}\}$. Compute the rank of $\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. \begin{proof} We first compute the values of $V_n$ for $0 \leq n \leq 3$ under the assumption the set of atoms $A$ at the bottom of the hierarchy is empty. \begin{align*} V_0 & = \emptyset \\ V_1 & = V_0 \cup \powerset{V_0} \\ & = \emptyset \cup \{\emptyset\} \\ & = \{\emptyset\} \\ V_2 & = V_1 \cup \powerset{V_1} \\ & = \{\emptyset\} \cup \powerset{\{\emptyset\}} \\ & = \{\emptyset\} \cup \{\emptyset, \{\emptyset\}\} \\ & = \{\emptyset, \{\emptyset\}\} \\ V_3 & = V_2 \cup \powerset{V_2} \\ & = \{\emptyset, \{\emptyset\}\} \cup \powerset{\{\emptyset, \{\emptyset\}\}} \\ & = \{\emptyset, \{\emptyset\}\} \cup \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\ & = \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \end{align*} It then immediately follows $\{\{\emptyset\}\}$ has rank $2$ and $\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$ has rank $3$. \end{proof} \subsection{\unverified{Exercise 1.6}}% \hyperlabel{sub:exercise-1.6} We have stated that $V_{\alpha + 1} = A \cup \powerset{V_\alpha}$. Prove this at least for $\alpha < 3$. \begin{proof} Let $A$ be the set of atoms in our set hierarchy. Let $P(n)$ be the predicate, "$V_{n + 1} = A \cup \powerset{V_n}$." We prove $P(n)$ holds true for all natural numbers $n \geq 1$ via induction. \paragraph{Base Case}% Let $n = 1$. By definition, $V_1 = V_0 \cup \powerset{V_0}$. By definition, $V_0 = A$. Therefore $V_1 = A \cup \powerset{V_0}$. This proves $P(1)$ holds true. \paragraph{Inductive Step}% Suppose $P(n)$ holds true for some $n \geq 1$. Consider $V_{n+1}$. By definition, $V_{n+1} = V_n \cup \powerset{V_n}$. Therefore, by the induction hypothesis, \begin{align} V_{n+1} & = V_n \cup \powerset{V_n} \nonumber \\ & = (A \cup \powerset{V_{n-1}}) \cup \powerset{V_n} \nonumber \\ & = A \cup (\powerset{V_{n-1}} \cup \powerset{V_n}) \hyperlabel{sub:exercise-1.6-eq1} \end{align} But $V_{n-1}$ is a subset of $V_n$. \nameref{sub:exercise-1.3} then implies $\powerset{V_{n-1}} \subseteq \powerset{V_n}$. This means \eqref{sub:exercise-1.6-eq1} can be simplified to $$V_{n+1} = A \cup \powerset{V_n},$$ proving $P(n+1)$ holds true. \paragraph{Conclusion}% By mathematical induction, it follows for all $n \geq 1$, $P(n)$ is true. \end{proof} \subsection{\unverified{Exercise 1.7}}% \hyperlabel{sub:exercise-1.7} List all the members of $V_3$. List all the members of $V_4$. (It is to be assumed here that there are no atoms.) \begin{proof} As seen in the proof of \nameref{sub:exercise-1.5}, $$V_3 = \{ \emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\} \}.$$ By \nameref{sub:exercise-1.6}, $V_4 = \powerset{V_3}$ (since it is assumed there are no atoms). Thus \begin{align*} & V_4 = \{ \\ & \qquad \emptyset, \\ & \qquad \{\emptyset\}, \\ & \qquad \{\{\emptyset\}\}, \\ & \qquad \{\{\{\emptyset\}\}\}, \\ & \qquad \{\{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\emptyset\}\}, \\ & \qquad \{\emptyset, \{\{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\{\emptyset\}, \{\{\emptyset\}\}\}, \\ & \qquad \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\ & \qquad \{\{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}, \\ & \qquad \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\} \\ & \}. \end{align*} \end{proof} \chapter{Axioms and Operations}% \hyperlabel{chap:axioms-operations} \section{Axioms}% \hyperlabel{sec:axioms} \subsection{\unverified{Theorem 2A}}% \hyperlabel{sub:theorem-2a} \begin{theorem}[2A] There is no set to which every set belongs. \end{theorem} \begin{note} This was revisited after reading Enderton's proof prior. \end{note} \begin{proof} Let $A$ be an arbitrary set. Define $B = \{ x \in A \mid x \not\in x \}$. By the \nameref{ref:subset-axioms}, $B$ is a set. Then $$B \in B \iff B \in A \land B \not\in B.$$ If $B \in A$, then $B \in B \iff B \not\in B$, a contradiction. Thus $B \not\in A$. Since this process holds for any set $A$, there must exist no set to which every set belongs. \end{proof} \subsection{\unverified{Theorem 2B}}% \hyperlabel{sub:theorem-2b} \begin{theorem}[2B] For any nonempty set $A$, there exists a unique set $B$ such that for any $x$, $$x \in B \iff x \text{ belongs to every member of } A.$$ \end{theorem} \begin{proof} Suppose $A$ is a nonempty set. This ensures the statement we are trying to prove does not vacuously hold for all sets $x$ (which would yield a contradiction due to \nameref{sub:theorem-2b}). By the \nameref{ref:union-axiom}, $\bigcup A$ is a set. Define $$B = \{ x \in \bigcup A \mid (\forall b \in A), x \in b \}.$$ By the \nameref{ref:subset-axioms}, $B$ is indeed a set. By construction, $$\forall x, x \in B \iff x \text{ belongs to every member of } A.$$ By the \nameref{ref:extensionality-axiom}, $B$ is unique. \end{proof} \section{Algebra of Sets}% \hyperlabel{sec:algebra-sets} \subsection{\verified{Commutative Laws}}% \hyperlabel{sub:commutative-laws} For any sets $A$ and $B$, \begin{enumerate}[(i)] \item $A \cup B = B \cup A$ \item $A \cap B = B \cap A$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.commutative\_law\_i} \lean{Mathlib/Data/Set/Basic}{Set.union\_comm} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.commutative\_law\_ii} \lean{Mathlib/Data/Set/Basic}{Set.inter\_comm} \begin{proof} Let $A$ and $B$ be sets. \paragraph{(i)}% By the definition of the union of sets, \begin{align*} A \cup B & = \{ x \mid x \in A \lor x \in B \} \\ & = \{ x \mid x \in B \lor x \in A \} \\ & = B \cup A. \end{align*} \paragraph{(ii)}% By the definition of the intersection of sets, \begin{align*} A \cap B & = \{ x \mid x \in A \land x \in B \} \\ & = \{ x \mid x \in B \land x \in A \} \\ & = B \land A. \end{align*} \end{proof} \subsection{\verified{Associative Laws}}% \hyperlabel{sub:associative-laws} For any sets $A$, $B$ and $C$, \begin{enumerate}[(i)] \item $A \cup (B \cup C) = (A \cup B) \cup C$ \item $A \cap (B \cap C) = (A \cap B) \cap C$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.associative\_law\_i} \lean{Mathlib/Data/Set/Basic}{Set.union\_assoc} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.associative\_law\_ii} \lean{Mathlib/Data/Set/Basic}{Set.inter\_assoc} \begin{proof} Let $A$, $B$, and $C$ be sets. \paragraph{(i)}% By the definition of the union of sets, \begin{align*} A \cup (B \cup C) & = \{ x \mid x \in A \lor x \in (B \cup C) \} \\ & = \{ x \mid x \in A \lor (x \in B \lor x \in C) \} \\ & = \{ x \mid (x \in A \lor x \in B) \lor x \in C \} \\ & = \{ x \mid x \in (A \cup B) \lor x \in C \} \\ & = (A \cup B) \cup C. \end{align*} \paragraph{(ii)}% By the definition of the intersection of sets, \begin{align*} A \cap (B \cap C) & = \{ x \mid x \in A \land x \in (B \cap C) \} \\ & = \{ x \mid x \in A \land (x \in B \land x \in C) \} \\ & = \{ x \mid (x \in A \land x \in B) \land x \in C \} \\ & = \{ x \mid x \in (A \cap B) \land x \in C \} \\ & = (A \cap B) \cap C. \end{align*} \end{proof} \subsection{\verified{Distributive Laws}}% \hyperlabel{sub:distributive-laws} For any sets $A$, $B$, and $C$, \begin{enumerate}[(i)] \item $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ \item $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.distributive\_law\_i} \lean{Mathlib/Data/Set/Basic}{Set.inter\_distrib\_left} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.distributive\_law\_ii} \lean{Mathlib/Data/Set/Basic}{Set.union\_distrib\_left} \begin{proof} Let $A$, $B$, and $C$ be sets. \paragraph{(i)}% By the definition of the union and intersection of sets, \begin{align*} A \cap (B \cup C) & = \{ x \mid x \in A \land x \in B \cup C \} \\ & = \{ x \mid x \in A \land (x \in B \lor x \in C) \} \\ & = \{ x \mid (x \in A \land x \in B) \lor (x \in A \land x \in C) \} \\ & = \{ x \mid x \in A \cap B \lor x \in A \cap C \} \\ & = (A \cap B) \cup (A \cap C). \end{align*} \paragraph{(ii)}% By the definition of the union and intersection of sets, \begin{align*} A \cup (B \cap C) & = \{ x \mid x \in A \lor x \in B \cap C \} \\ & = \{ x \mid x \in A \lor (x \in B \land x \in C) \} \\ & = \{ x \mid (x \in A \lor x \in B) \land (x \in A \lor x \in C) \} \\ & = \{ x \mid x \in A \cup B \land x \in A \cup C \} \\ & = (A \cup B) \cap (A \cup C). \end{align*} \end{proof} \subsection{\verified{De Morgan's Laws}}% \hyperlabel{sub:de-morgans-laws} For any sets $A$, $B$, and $C$, \begin{enumerate}[(i)] \item $C - (A \cup B) = (C - A) \cap (C - B)$ \item $C - (A \cap B) = (C - A) \cup (C - B)$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.de\_morgans\_law\_i} \lean{Mathlib/Data/Set/Basic}{Set.diff\_inter\_diff} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.de\_morgans\_law\_ii} \lean{Mathlib/Data/Set/Basic}{Set.diff\_inter} \begin{proof} Let $A$, $B$, and $C$ be sets. \paragraph{(i)}% By definition of the union, intersection, and relative complements of sets, \begin{align*} C - (A \cup B) & = \{ x \mid x \in C \land x \not\in A \cup B \} \\ & = \{ x \mid x \in C \land \neg(x \in A \lor x \in B) \} \\ & = \{ x \mid x \in C \land (x \not\in A \land x \not\in B) \} \\ & = \{ x \mid (x \in C \land x \not\in A) \land (x \in C \land x \not\in B) \} \\ & = \{ x \mid x \in (C - A) \land x \in (C - B) \} \\ & = (C - A) \cap (C - B). \end{align*} \paragraph{(ii)}% By definition of the union, intersection, and relative complements of sets, \begin{align*} C - (A \cap B) & = \{ x \mid x \in C \land x \not\in A \cap B \} \\ & = \{ x \mid x \in C \land \neg(x \in A \land x \in B) \} \\ & = \{ x \mid x \in C \land (x \not\in A \lor x \not\in B) \} \\ & = \{ x \mid (x \in C \land x \not\in A) \lor (x \in C \land x \not\in B) \} \\ & = \{ x \mid x \in C - A \lor x \in C - B \} \\ & = (C - A) \cup (C - B). \end{align*} \end{proof} \subsection{\verified{% Identities Involving \texorpdfstring{$\emptyset$}{the Empty Set}}}% \hyperlabel{sub:identitives-involving-empty-set} For any set $A$, \begin{enumerate}[(i)] \item $A \cup \emptyset = A$ \item $A \cap \emptyset = \emptyset$ \item $A \cap (C - A) = \emptyset$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.emptyset\_identity\_i} \lean{Mathlib/Data/Set/Basic}{Set.union\_empty} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.emptyset\_identity\_ii} \lean{Mathlib/Data/Set/Basic}{Set.inter\_empty} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.emptyset\_identity\_iii} \lean{Mathlib/Data/Set/Basic}{Set.inter\_diff\_self} \begin{proof} Let $A$ be an arbitrary set. \paragraph{(i)}% By definition of the emptyset and union of sets, \begin{align*} A \cup \emptyset & = \{ x \mid x \in A \lor x \in \emptyset \} \\ & = \{ x \mid x \in A \lor F \} \\ & = \{ x \mid x \in A \} \\ & = A. \end{align*} \paragraph{(ii)}% By definition of the emptyset and intersection of sets, \begin{align*} A \cap \emptyset & = \{ x \mid x \in A \land x \in \emptyset \} \\ & = \{ x \mid x \in A \land F \} \\ & = \{ x \mid F \} \\ & = \emptyset. \end{align*} \paragraph{(iii)}% By definition of the emptyset, and the intersection and relative complement of sets, \begin{align*} A \cap (C - A) & = \{ x \mid x \in A \land x \in C - A \} \\ & = \{ x \mid x \in A \land (x \in C \land x \not\in A) \} \\ & = \{ x \mid x \in C \land F \} \\ & = \{ x \mid F \} \\ & = \emptyset. \end{align*} \end{proof} \subsection{\verified{Monotonicity}}% \hyperlabel{sub:monotonicity} For any sets $A$, $B$, and $C$, \begin{enumerate}[(i)] \item $A \subseteq B \Rightarrow A \cup C \subseteq B \cup C$ \item $A \subseteq B \Rightarrow A \cap C \subseteq B \cap C$ \item $A \subseteq B \Rightarrow \bigcup A \subseteq \bigcup B$ \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.monotonicity\_i} \lean{Mathlib/Data/Set/Basic} {Set.union\_subset\_union\_left} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.monotonicity\_ii} \lean{Mathlib/Data/Set/Basic} {Set.inter\_subset\_inter\_left} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.monotonicity\_iii} \lean{Mathlib/Data/Set/Lattice} {Set.sUnion\_mono} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. \paragraph{(i)}% Suppose $A \subseteq B$. Let $x \in A \cup C$. There are two cases to consider. \subparagraph{Case 1}% Suppose $x \in A$. Then, by definition of the subset, $x \in B$. Therefore $x \in B \cup C$. \subparagraph{Case 2}% Suppose $x \in C$. Then $x$ is trivially a member of $B \cup C$. \subparagraph{Conclusion}% Since these cases are exhaustive and both imply $x \in B \cup C$, it follows $A \cup C \subseteq B \cup C$. \paragraph{(ii)}% Suppose $A \subseteq B$. Let $x \in A \cap C$. Then, by definition of the intersection of sets, $x \in A$ and $x \in C$. By definition of the subset, $x \in A$ implies $x \in B$. Therefore $x \in B$ and $x \in C$. That is, $x \in B \cap C$. Since this holds for arbitrary $x \in A \cap C$, it follows $A \cap C \subseteq B \cap C$. \paragraph{(iii)}% Suppose $A \subseteq B$. Let $x \in \bigcup A$. Then, by definition of the union of sets, there exists some $b \in A$ such that $x \in b$. By definition of the subset, $b \in B$ as well. Another application of the definition of the union of sets immediately implies that $x$ is a member of $\bigcup B$. \end{proof} \subsection{\verified{Anti-monotonicity}}% \hyperlabel{sub:anti-monotonicity} For any sets $A$, $B$, and $C$, \begin{enumerate}[(i)] \item $A \subseteq B \Rightarrow C - B \subseteq C - A$ \item $\emptyset \neq A \subseteq B \Rightarrow \bigcap B \subseteq \bigcap A$. \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.anti\_monotonicity\_i} \lean{Mathlib/Data/Set/Basic} {Set.diff\_subset\_diff\_right} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.anti\_monotonicity\_ii} \lean{Mathlib/Data/Set/Lattice} {Set.sInter\_subset\_sInter} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. \paragraph{(i)}% Suppose $A \subseteq B$. Let $x \in C - B$. By definition of the relative complement, $x \in C$ and $x \not\in B$. Then $x$ cannot be a member of $A$, since otherwise this would contradict our subset hypothesis. That is, $x \in C$ and $x \not\in A$. Therefore $x \in C - A$. Since this holds for arbitrary $x \in C - B$, it follows that $C - B \subseteq C - A$. \paragraph{(ii)}% Suppose $A \neq \emptyset$ and $A \subseteq B$. Then $B \neq \emptyset$. Let $x \in \bigcap B$. By definition of the intersection of sets, for all $b \in B$, $x \in b$. But then, by definition of the subset, for all $a \in A$, $x \in a$. Therefore $x \in \bigcap A$. Since this holds for arbitrary $x \in \bigcap B$, it follows that $\bigcap B \subseteq \bigcap A$. \end{proof} \subsection{\unverified{General Distributive Laws}}% \hyperlabel{sub:general-distributive-laws} For any sets $A$ and $\mathscr{B}$, \begin{enumerate}[(i)] \item $A \cup \bigcap \mathscr{B} = \bigcap\; \{ A \cup X \mid X \in \mathscr{B} \} \quad\text{for}\quad \mathscr{B} \neq \emptyset$ \item $A \cap \bigcup \mathscr{B} = \bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}$ \end{enumerate} \begin{proof} Let $A$ and $\mathscr{B}$ be sets. \paragraph{(i)}% Suppose $\mathscr{B}$ is nonempty. Then $\bigcap \mathscr{B}$ is defined. By definition of the union and intersection of sets, \begin{align*} A \cup \bigcap \mathscr{B} & = \{ x \mid x \in A \lor x \in \bigcap \mathscr{B} \} \\ & = \{ x \mid x \in A \lor x \in \{ y \mid (\forall b \in \mathscr{B}), y \in b \}\} \\ & = \{ x \mid x \in A \lor (\forall b \in \mathscr{B}), x \in b \} \\ & = \{ x \mid \forall b \in \mathscr{B}, x \in A \lor x \in b \} \\ & = \{ x \mid \forall b \in \mathscr{B}, x \in A \cup b \} \\ & = \{ x \mid x \in \bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}\} \\ & = \bigcap\; \{ A \cup X \mid X \in \mathscr{B} \}. \end{align*} \paragraph{(ii)}% By definition of the intersection and union of sets, \begin{align*} A \cap \bigcup \mathscr{B} & = \{ x \mid x \in A \land x \in \bigcup \mathscr{B} \} \\ & = \{ x \mid x \in A \land x \in \{ y \mid (\exists b \in \mathscr{B}), y \in b \}\} \\ & = \{ x \mid x \in A \land (\exists b \in \mathscr{B}), x \in b \} \\ & = \{ x \mid \exists b \in \mathscr{B}, x \in A \land x \in b \} \\ & = \{ x \mid \exists b \in \mathscr{B} x \in A \cap b \} \\ & = \{ x \mid x \in \bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}\} \\ & = \bigcup\; \{ A \cap X \mid X \in \mathscr{B} \}. \end{align*} \end{proof} \subsection{\unverified{General De Morgan's Laws}}% \hyperlabel{sub:general-de-morgans-laws} For any set $C$ and $\mathscr{A} \neq \emptyset$, \begin{enumerate}[(i)] \item $C - \bigcup \mathscr{A} = \bigcap\; \{ C - X \mid X \in \mathscr{A} \}$ \item $C - \bigcap \mathscr{A} = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}$ \end{enumerate} \begin{proof} Let $C$ and $\mathscr{A}$ be sets such that $\mathscr{A} \neq \emptyset$. \paragraph{(i)}% By definition of the relative complement, union, and intersection of sets, \begin{align*} C - \bigcup \mathscr{A} & = \{ x \mid x \in C \land x \not\in \bigcup \mathscr{A} \} \\ & = \{ x \mid x \in C \land x \not\in \{ y \mid (\exists b \in \mathscr{A}) y \in b \}\} \\ & = \{ x \mid x \in C \land \neg(\exists b \in \mathscr{A}, x \in b) \} \\ & = \{ x \mid x \in C \land (\forall b \in \mathscr{A}, x \not\in b) \} \\ & = \{ x \mid \forall b \in \mathscr{A}, x \in C \land x \not\in b \} \\ & = \{ x \mid \forall b \in \mathscr{A}, x \in C - b \} \\ & = \{ x \mid x \in \bigcap\; \{ C - X \mid X \in \mathscr{A} \} \\ & = \bigcap\; \{ C - X \mid X \in \mathscr{A} \}. \end{align*} \paragraph{(ii)}% By definition of the relative complement, union, and intersection of sets, \begin{align*} C - \bigcap \mathscr{A} & = \{ x \mid x \in C \land x \not\in \bigcap \mathscr{A} \} \\ & = \{ x \mid x \in C \land x \not\in \{ y \mid (\forall b \in \mathscr{A}) y \in b \}\} \\ & = \{ x \mid x \in C \land \neg(\forall b \in \mathscr{A}, x \in b) \} \\ & = \{ x \mid x \in C \land \exists b \in \mathscr{A}, x \not\in b \} \\ & = \{ x \mid \exists b \in \mathscr{A}, x \in C \land x \not\in b \} \\ & = \{ x \mid \exists b \in \mathscr{A}, x \in C - b \} \\ & = \{ x \mid x \in \bigcup\; \{ C - X \mid X \in \mathscr{A} \} \} \\ & = \bigcup\; \{ C - X \mid X \in \mathscr{A} \}. \end{align*} \end{proof} \subsection{\verified{% \texorpdfstring{$\cap$/$-$}{Intersection/Difference} Associativity}}% \hyperlabel{sub:intersection-difference-associativity} Let $A$, $B$, and $C$ be sets. Then $A \cap (B - C) = (A \cap B) - C$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.inter\_diff\_assoc} \lean{Mathlib/Data/Set/Basic} {Set.inter\_diff\_assoc} \begin{proof} Let $A$, $B$, and $C$ be sets. By definition of the intersection and relative complement of sets, \begin{align*} A \cap (B - C) & = \{ x \mid x \in A \land x \in B - C \} \\ & = \{ x \mid x \in A \land (x \in B \land x \not\in C) \} \\ & = \{ x \mid (x \in A \land x \in B) \land x \not\in C \} \\ & = \{ x \mid x \in A \cap B \land x \not \in C \} \\ & = (A \cap B) - C. \end{align*} \end{proof} \subsection{\verified{Nonmembership of Symmetric Difference}} \hyperlabel{sub:nonmembership-symmetric-difference} Let $A$ and $B$ be sets. $x \not\in A + B$ if and only if either $x \in A \cap B$ or $x \not\in A \cup B$. \code*{Common/Set/Basic} {Set.not\_mem\_symm\_diff\_inter\_or\_not\_union} \begin{proof} By definition of the \nameref{ref:symmetric-difference}, \begin{align*} x \not\in A + B & = \neg(x \in A + B) \\ & = \neg[x \in (A - B) \cup (B - A)] \\ & = \neg[x \in (A - B) \lor x \in (B - A)] \\ & = \neg[(x \in A \land x \not\in B) \lor (x \in B \land x \not\in A)] \\ & = \neg(x \in A \land x \not\in B) \land \neg(x \in B \land x \not\in A) \\ & = (x \not\in A \lor x \in B) \land (x \not\in B \lor x \in A) \\ & = ((x \not\in A \lor x \in B) \land x \not\in B) \lor ((x \not\in A \lor x \in B) \land x \in A) \\ & = (x \not\in A \land x \not\in B) \lor (x \in B \land x \in A) \\ & = \neg(x \in A \lor x \in B) \lor (x \in B \land x \in A) \\ & = x \not\in A \cup B \text{ or } x \in A \cap B. \end{align*} \end{proof} \section{Exercises 2}% \hyperlabel{sec:exercises-2} \subsection{\verified{Exercise 2.1}}% \hyperlabel{sub:exercise-2.1} Assume that $A$ is the set of integers divisible by $4$. Similarly assume that $B$ and $C$ are the sets of integers divisible by $9$ and $10$, respectively. What is in $A \cap B \cap C$? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_1} \begin{answer} The set of integers divisible by $4$, $9$, and $10$. \end{answer} \subsection{\verified{Exercise 2.2}}% \hyperlabel{sub:exercise-2.2} Give an example of sets $A$ and $B$ for which $\bigcup A = \bigcup B$ but $A \neq B$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_2} \begin{answer} Let $A = \{\{1\}, \{2\}\}$ and $B = \{\{1, 2\}\}$. \end{answer} \subsection{\verified{Exercise 2.3}}% \hyperlabel{sub:exercise-2.3} Show that every member of a set $A$ is a subset of $\bigcup A$. (This was stated as an example in this section.) \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_3} \begin{proof} Let $x \in A$. By definition, $$\bigcup A = \{ y \mid (\exists b \in A) y \in b\}.$$ Then $\{ y \mid y \in x\} \subseteq \bigcup A$. But $\{ y \mid y \in x\} = x$. Thus $x \subseteq \bigcup A$. \end{proof} \subsection{\verified{Exercise 2.4}}% \hyperlabel{sub:exercise-2.4} Show that if $A \subseteq B$, then $\bigcup A \subseteq \bigcup B$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_4} \begin{proof} Let $A$ and $B$ be sets such that $A \subseteq B$. Let $x \in \bigcup A$. By definition of the union, there exists some $b \in A$ such that $x \in b$. By definition of the subset, $b \in B$. This immediatley implies $x \in \bigcup B$. Since this holds for all $x \in \bigcup A$, it follows $\bigcup A \subseteq \bigcup B$. \end{proof} \subsection{\verified{Exercise 2.5}}% \hyperlabel{sub:exercise-2.5} Assume that every member of $\mathscr{A}$ is a subset of $B$. Show that $\bigcup \mathscr{A} \subseteq B$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_5} \begin{proof} Let $x \in \bigcup \mathscr{A}$. By definition, $$\bigcup \mathscr{A} = \{ y \mid (\exists b \in A)y \in b \}.$$ Then there exists some $b \in A$ such that $x \in b$. By hypothesis, $b \subseteq B$. Thus $x$ must also be a member of $B$. Since this holds for all $x \in \bigcup \mathscr{A}$, it follows $\bigcup \mathscr{A} \subseteq B$. \end{proof} \subsection{\verified{Exercise 2.6a}}% \hyperlabel{sub:exercise-2.6a} Show that for any set $A$, $\bigcup \powerset{A} = A$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_6a} \begin{proof} We prove that (i) $\bigcup \powerset{A} \subseteq A$ and (ii) $A \subseteq \bigcup \powerset{A}$. \paragraph{(i)}% \hyperlabel{par:exercise-2.6a-i} By definition, the \nameref{ref:power-set} of $A$ is the set of all subsets of $A$. In other words, every member of $\powerset{A}$ is a subset of $A$. By \nameref{sub:exercise-2.5}, $\bigcup \powerset{A} \subseteq A$. \paragraph{(ii)}% \hyperlabel{par:exercise-2.6a-ii} Let $x \in A$. By definition of the power set of $A$, $\{x\} \in \powerset{A}$. By definition of the union, $$\bigcup \powerset{A} = \{ y \mid (\exists b \in \powerset{A}), y \in b).$$ Since $x \in \{x\}$ and $\{x\} \in \powerset{A}$, it follows $x \in \bigcup \powerset{A}$. Thus $A \subseteq \bigcup \powerset{A}$. \paragraph{Conclusion}% By \nameref{par:exercise-2.6a-i} and \nameref{par:exercise-2.6a-ii}, $\bigcup \powerset{A} = A$. \end{proof} \subsection{\verified{Exercise 2.6b}}% \hyperlabel{sub:exercise-2.6b} Show that $A \subseteq \powerset{\bigcup A}$. Under what conditions does equality hold? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_6b} \begin{proof} Let $x \in A$. By \nameref{sub:exercise-2.3}, $x$ is a subset of $\bigcup A$. By the definition of the \nameref{ref:power-set}, $$\powerset{\bigcup A} = \{ y \mid y \subseteq \bigcup A \}.$$ Therefore $x \in \powerset{\bigcup A}$. Since this holds for all $x \in A$, $A \subseteq \powerset{\bigcup A}$. \suitdivider We show equality holds if and only if there exists some set $B$ such that $A = \powerset{B}$. \paragraph{($\Rightarrow$)}% \hyperlabel{par:exercise-2.6b-right} Suppose $A = \powerset{\bigcup A}$. Then our statement immediately follows by settings $B = \bigcup A$. \paragraph{($\Leftarrow$)}% \hyperlabel{par:exercise-2.6b-left} Suppose there exists some set $B$ such that $A = \powerset{B}$. Therefore \begin{align*} \powerset{\bigcup A} & = \powerset{\left(\bigcup {\powerset {B}}\right)} \\ & = \powerset{B} & \textref{sub:exercise-2.6a} \\ & = A. \end{align*} \paragraph{Conclusion}% By \nameref{par:exercise-2.6b-right} and \nameref{par:exercise-2.6b-left}, $A = \powerset{\bigcup A}$ if and only if there exists some set $B$ such that $A = \powerset{B}$. \end{proof} \subsection{\verified{Exercise 2.7a}}% \hyperlabel{sub:exercise-2.7a} Show that for any sets $A$ and $B$, $$\powerset{A} \cap \powerset{B} = \powerset{(A \cap B)}.$$ \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_7a} \begin{proof} Let $A$ and $B$ be arbitrary sets. We show that $\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}$ and then show that $\powerset{A} \cap \powerset{B} \supseteq \powerset{(A \cap B)}$. \paragraph{($\subseteq$)}% Let $x \in \powerset{A} \cap \powerset{B}$. That is, $x \in \powerset{A}$ and $x \in \powerset{B}$. By the definition of the \nameref{ref:power-set}, \begin{align*} \powerset{A} & = \{ y \mid y \subseteq A \} \\ \powerset{B} & = \{ y \mid y \subseteq B \} \end{align*} Thus $x \subseteq A$ and $x \subseteq B$, meaning $x \subseteq A \cap B$. But then $x \in \powerset{(A \cap B)}$, the set of all subsets of $A \cap B$. Since this holds for all $x \in \powerset{A} \cap \powerset{B}$, it follows $$\powerset{A} \cap \powerset{B} \subseteq \powerset{(A \cap B)}.$$ \paragraph{($\supseteq$)}% Let $x \in \powerset{(A \cap B)}$. By the definition of the \nameref{ref:power-set}, $$\powerset{(A \cap B)} = \{ y \mid y \subseteq A \cap B \}.$$ Thus $x \subseteq A \cap B$, meaning $x \subseteq A$ and $x \subseteq B$. But this implies $x \in \powerset{A}$, the set of all subsets of $A$. Likewise $x \in \powerset{B}$, the set of all subsets of $B$. Thus $x \in \powerset{A} \cap \powerset{B}$. Since this holds for all $x \in \powerset{(A \cap B)}$, it follows $$\powerset{(A \cap B)} \subseteq \powerset{A} \cap \powerset{B}.$$ \paragraph{Conclusion}% Since each side of our identity is a subset of the other, $$\powerset{(A \cap B)} = \powerset{A} \cap \powerset{B}.$$ \end{proof} \subsection{\verified{Exercise 2.7b}}% \hyperlabel{sub:exercise-2.7b} Show that $\powerset{A} \cup \powerset{B} \subseteq \powerset{(A \cup B)}$. Under what conditions does equality hold? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_7b\_i} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_7b\_ii} \begin{proof} Let $x \in \powerset{A} \cup \powerset{B}$. By definition, $x \in \powerset{A}$ or $x \in \powerset{B}$ (or both). By the definition of the \nameref{ref:power-set}, \begin{align*} \powerset{A} &= \{ y \mid y \subseteq A \} \\ \powerset{B} &= \{ y \mid y \subseteq B \}. \end{align*} Thus $x \subseteq A$ or $x \subseteq B$. Therefore $x \subseteq A \cup B$. But then $x \in \powerset{(A \cup B)}$, the set of all subsets of $A \cup B$. \suitdivider We show equality holds if and only if one of $A$ or $B$ is a subset of the other. \paragraph{($\Rightarrow$)}% \hyperlabel{par:exercise-2.7b-right} Suppose \begin{equation} \hyperlabel{sub:exercise-2.7b-eq1} \powerset{A} \cup \powerset{B} = \powerset{(A \cup B)}. \end{equation} By the definition of the \nameref{ref:power-set}, $A \cup B \in \powerset{(A \cup B)}$. Then \eqref{sub:exercise-2.7b-eq1} implies $A \cup B \in \powerset{A} \cup \powerset{B}$. That is, $A \cup B \in \powerset{A}$ or $A \cup B \in \powerset{B}$ (or both). For the sake of contradiction, suppose $A \not\subseteq B$ and $B \not\subseteq A$. Then there exists an element $x \in A$ such that $x \not\in B$ and there exists an element $y \in B$ such that $y \not\in A$. But then $A \cup B \not\in \powerset{A}$ since $y$ cannot be a member of a member of $\powerset{A}$. Likewise, $A \cup B \not\in \powerset{B}$ since $x$ cannot be a member of a member of $\powerset{B}$. Therefore our assumption is incorrect. In other words, $A \subseteq B$ or $B \subseteq A$. \paragraph{($\Leftarrow$)}% \hyperlabel{par:exercise-2.7b-left} WLOG, suppose $A \subseteq B$. Then, by \nameref{sub:exercise-1.3}, $\powerset{A} \subseteq \powerset{B}$. Thus \begin{align*} \powerset{A} \cup \powerset{B} & = \powerset{B} \\ & = \powerset{A \cup B}. \end{align*} \paragraph{Conclusion}% By \nameref{par:exercise-2.7b-right} and \nameref{par:exercise-2.7b-left}, it follows $\powerset{A} \cup \powerset{B} \subseteq \powerset{(A \cup B)}$ if and only if $A \subseteq B$ or $B \subseteq A$. \end{proof} \subsection{\unverified{Exercise 2.8}}% \hyperlabel{sub:exercise-2.8} Show that there is no set to which every singleton (that is, every set of the form $\{x\}$) belongs. [\textit{Suggestion}: Show that from such a set, we could construct a set to which every set belonged.] \begin{proof} We proceed by contradiction. Suppose there existed a set $A$ consisting of every singleton. Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set. But this "set" is precisely the class of all sets, which is \textit{not} a set. Thus our original assumption was incorrect. That is, there is no set to which every singleton belongs. \end{proof} \subsection{\verified{Exercise 2.9}}% \hyperlabel{sub:exercise-2.9} Give an example of sets $a$ and $B$ for which $a \in B$ but $\powerset{a} \not\in \powerset{B}$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_9} \begin{answer} Let $a = \{1\}$ and $B = \{\{1\}\}$. Then \begin{align*} \powerset{a} & = \{\emptyset, \{1\}\} \\ \powerset{B} & = \{\emptyset, \{\{1\}\}\}. \end{align*} It immediately follows that $\powerset{a} \not\in \powerset{B}$. \end{answer} \subsection{\verified{Exercise 2.10}}% \hyperlabel{sub:exercise-2.10} Show that if $a \in B$, then $\powerset{a} \in \powerset{\powerset{\bigcup B}}$. [\textit{Suggestion}: If you need help, look in the Appendix.] \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_10} \begin{proof} Suppose $a \in B$. By \nameref{sub:exercise-2.3}, $a \subseteq \bigcup B$. By \nameref{sub:exercise-1.3}, $\powerset{a} \subseteq \powerset{\bigcup B}$. By the definition of the \nameref{ref:power-set}, $$\powerset{\powerset{\bigcup B}} = \{ y \mid y \subseteq \powerset{\bigcup B} \}.$$ Therefore $\powerset{a} \in \powerset{\powerset{\bigcup B}}$. \end{proof} \subsection{\verified{Exercise 2.11}}% \hyperlabel{sub:exercise-2.11} Show that for any sets $A$ and $B$, $$A = (A \cap B) \cup (A - B) \quad\text{and}\quad A \cup (B - A) = A \cup B.$$ \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_11\_i} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_11\_ii} \begin{proof} Let $A$ and $B$ be sets. We prove that \begin{enumerate}[(i)] \item $A = (A \cap B) \cup (A - B)$ \item $A \cup (B - A) = A \cup B$ \end{enumerate} \paragraph{(i)}% By definition of the intersection, union, and relative complements of sets, \begin{align*} (A \cap B) \cup (A - B) & = \{ x \mid x \in A \cap B \lor x \in A - B \} \\ & = \{ x \mid x \in \{ y \mid y \in A \land y \in B \} \lor x \in A - B \} \\ & = \{ x \mid (x \in A \land x \in B) \lor x \in A - B \} \\ & = \{ x \mid (x \in A \land x \in B) \lor x \in \{ y \mid y \in A \land y \not\in B \} \} \\ & = \{ x \mid (x \in A \land x \in B) \lor (x \in A \land x \not\in B) \} \\ & = \{ x \mid x \in A \lor (x \in B \land x \not\in B) \} \\ & = \{ x \mid x \in A \lor F \} \\ & = \{ x \mid x \in A \} \\ & = A. \end{align*} \paragraph{(ii)}% By definition of the union and relative complements of sets, \begin{align*} A \cup (B - A) & = \{ x \mid x \in A \lor x \in B - A \} \\ & = \{ x \mid x \in A \lor x \in \{ y \mid y \in B \land y \not\in A \} \} \\ & = \{ x \mid x \in A \lor (x \in B \land x \not\in A) \} \\ & = \{ x \mid (x \in A \lor x \in B) \land (x \in A \lor x \not\in A) \} \\ & = \{ x \mid (x \in A \lor x \in B) \land T \} \\ & = \{ x \mid x \in A \lor x \in B \} \\ & = \{ x \mid x \in A \cup B \} \\ & = A \cup B. \end{align*} \end{proof} \subsection{\verified{Exercise 2.12}}% \hyperlabel{sub:exercise-2.12} Verify the following identity (one of De Morgan's laws): $$C - (A \cap B) = (C - A) \cup (C - B).$$ \begin{proof} Refer to \nameref{sub:de-morgans-laws}. \end{proof} \subsection{\verified{Exercise 2.13}}% \hyperlabel{sub:exercise-2.13} Show that if $A \subseteq B$, then $C - B \subseteq C - A$. \begin{proof} Refer to \nameref{sub:anti-monotonicity}. \end{proof} \subsection{\verified{Exercise 2.14}}% \hyperlabel{sub:exercise-2.14} Show by example that for some sets $A$, $B$, and $C$, the set $A - (B - C)$ is different from $(A - B) - C$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_14} \begin{proof} Let $A = \{1, 2, 3\}$, $B = \{2, 3, 4\}$, and $C = \{3, 4, 5\}$. Then \begin{align*} A - (B - C) & = \{1, 2, 3\} - (\{2, 3, 4\} - \{3, 4, 5\}) \\ & = \{1, 2, 3\} - \{2\} \\ & = \{1, 3\} \end{align*} but \begin{align*} (A - B) - C & = (\{1, 2, 3\} - \{2, 3, 4\}) - \{3, 4, 5\} \\ & = \{1\} - \{3, 4, 5\} \\ & = \{1\}. \end{align*} \end{proof} \subsection{\verified{Exercise 2.15a}}% \hyperlabel{sub:exercise-2.15a} Show that $A \cap (B + C) = (A \cap B) + (A \cap C)$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_15a} \lean{Mathlib/Data/Set/Basic} {Set.inter\_symmDiff\_distrib\_left} \begin{proof} By definition of the intersection, \nameref{ref:symmetric-difference}, and relative complement of sets, \begin{align*} (A & \cap B) + (A \cap C) \\ & = [(A \cap B) - (A \cap C)] \cup [(A \cap C) - (A \cap B)] \\ & = [(A \cap B) - A] \\ & \qquad \cup [(A \cap B) - C] \\ & \qquad \cup [(A \cap C) - A] \\ & \qquad \cup [(A \cap C) - B] & \textref{sub:de-morgans-laws} \\ & = [A \cap (B - A)] \\ & \qquad \cup [A \cap (B - C)] \\ & \qquad \cup [A \cap (C - A)] \\ & \qquad \cup [A \cap (C - B)] & \textref{sub:intersection-difference-associativity} \\ & = \emptyset \\ & \qquad \cup [A \cap (B - C)] \\ & \qquad \cup \emptyset \\ & \qquad \cup [A \cap (C - B)] & \textref{sub:identitives-involving-empty-set} \\ & = [A \cap (B - C)] \cup [A \cap (C - B)] \\ & = A \cap [(B - C) \cup (C - B)] & \textref{sub:distributive-laws} \\ & = A \cap (B + C). \end{align*} \end{proof} \subsection{\verified{Exercise 2.15b}}% \hyperlabel{sub:exercise-2.15b} Show that $A + (B + C) = (A + B) + C$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_15b} \lean{Mathlib/Order/SymmDiff}{symmDiff\_assoc} \begin{proof} Let $A$, $B$, and $C$ be sets. We prove that \begin{enumerate}[(i)] \item $A + (B + C) \subseteq (A + B) + C$ \item $(A + B) + C \subseteq A + (B + C)$ \end{enumerate} \paragraph{(i)}% \hyperlabel{par:exercise-2.15b-i} Let $x \in A + (B + C)$. Then $x$ is in $A$ or in $B + C$, but not both. There are two cases to consider: \subparagraph{Case 1}% Suppose $x \in A$ and $x \not\in B + C$. Then, by \nameref{sub:nonmembership-symmetric-difference}, (a) $x \in B \cap C$ or (b) $x \not\in B \cup C$. Suppose (a) was true. That is, $x \in B$ and $x \in C$. Since $x$ is a member of $A$ and $B$, $x \not\in (A + B)$. Since $x$ is not a member of $A + B$ but is a member of $C$, $x \in (A + B) + C$. Now suppose (b) was true. That is, $x \not\in B$ and $x \not\in C$. Since $x$ is a member of $A$ but not $B$, $x \in (A + B)$. Since $x$ is a member of $A + B$ but not $C$, $x \in (A + B) + C$. \subparagraph{Case 2}% Suppose $x \in B + C$ and $x \not\in A$. Then (a) $x \in B$ or (b) $x \in C$ but not both. Suppose (a) was true. That is, $x \in B$ and $x \not\in C$. Since $x$ is not a member of $A$ and is a member of $B$, $x \in A + B$. Since $x$ is a member of $A + B$ but not $C$, $x \in (A + B) + C$. Now suppose (b) was true. That is, $x \not\in B$ and $x \in C$. Since $x$ is not a member of $A$ nor a member of $B$, $x \not\in A + B$. Since $x$ is not a member of $A + B$ but is a member of $C$, $x \in (A + B) + C$. \paragraph{(ii)}% \hyperlabel{par:exercise-2.15b-ii} Let $x \in (A + B) + C$. Then $x$ is in $A + B$ or in $C$, but not both. There are two cases to consider: \subparagraph{Case 1}% Suppose $x \in A + B$ and $x \not\in C$. Then (a) $x \in A$ or (b) $x \in B$ but not both. Suppose (a) was true. That is, $x \in A$ and $x \not\in B$. Since $x$ is not a member of $B$ nor $C$, $x \not\in B + C$. Since $x$ is not a member of $B + C$ but is a member of $A$, $x \in A + (B + C)$. Now Suppose (b) was true. That is, $x \not\in A$ and $x \in B$. Since $x$ is a member of $B$ and not of $C$, then $x \in B + C$. Since $x$ is a member of $B + C$ and not of $A$, $x \in A + (B + C)$. \subparagraph{Case 2}% Suppose $x \not\in A + B$ and $x \in C$. Then, by \nameref{sub:nonmembership-symmetric-difference}, (a) $x \in A \cap B$ or (b) $x \not\in A \cup B$. Suppose (a) was true. That is, $x \in A \land x \in B$. Since $x$ is a member of $B$ and $C$, $x \not\in B + C$. Since $x$ is not a member of $B + C$ but is a member of $A$, $x \in A + (B + C)$. Now suppose (b) was true. That is, $x \not\in A$ and $x \not\in B$. Since $x$ is not a member of $B$ but is a member of $C$, $x \in B + C$. Since $x$ is a member of $B + C$ but not of $A$, $x \in A + (B + C)$. \paragraph{Conclusion}% In both \nameref{par:exercise-2.15b-i} and \nameref{par:exercise-2.15b-ii}, the subcases are exhaustive and prove the desired subset relation. Therefore $A + (B + C) = (A + B) + C$. \end{proof} \subsection{\verified{Exercise 2.16}}% \hyperlabel{sub:exercise-2.16} Simplify: $$[(A \cup B \cup C) \cap (A \cup B)] - [(A \cup (B - C)) \cap A].$$ \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_16} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. Then \begin{align*} [(A \cup B \cup C) \cap (A \cup B)] & - [(A \cup (B - C)) \cap A] \\ & = [A \cup B] - [A] \\ & = \{ x \mid x \in (A \cup B) \land x \not\in A \} \\ & = \{ x \mid x \in \{ y \mid y \in A \lor y \in B \} \land x \not\in A \} \\ & = \{ x \mid (x \in A \lor x \in B) \land x \not\in A \} \\\ & = \{ x \mid (x \in A \land x \not\in A) \lor (x \in B \land x \not\in A) \} \\ & = \{ x \mid F \lor (X \in B \land x \not\in A) \} \\ & = \{ x \mid x \in B \land x \not\in A \} \\ & = B - A. \end{align*} \end{proof} \subsection{\verified{Exercise 2.17}}% \hyperlabel{sub:exercise-2.17} Show that the following four conditions are equivalent. \begin{enumerate}[(a)] \item $A \subseteq B$, \item $A - B = \emptyset$, \item $A \cup B = B$, \item $A \cap B = A$. \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_17\_i} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_17\_ii} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_17\_iii} \code{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_17\_iv} \begin{proof} Let $A$ and $B$ be arbitrary sets. We show that (i) $(a) \Rightarrow (b)$, (ii) $(b) \Rightarrow (c)$, (iii) $(c) \Rightarrow (d)$, and (iv) $(d) \Rightarrow (a)$. \paragraph{(i)}% Suppose $A \subseteq B$. That is, $\forall t, t \in A \Rightarrow t \in B$. Then there is no element such that $t \in A$ and $t \not\in B$. By definition of the relative complement, this immediately implies $A - B = \emptyset$. \paragraph{(ii)}% Suppose $A - B = \emptyset$. By definition of the relative complement, $$A - B = \emptyset = \{ x \mid x \in A \land x \not\in B \}.$$ Then, for all $t$, $\neg(t \in A \land t \not\in B) = t \not\in A \lor t \in B$. This implies, by definition of the subset, that $A \subseteq B$. It then immediately follows that $A \cup B = B$. \paragraph{(iii)}% Suppose $A \cup B = B$. Then there is no member of $A$ that is not a member of $B$. In other words, $A \subseteq B$. This immediately implies $A \cap B = A$. \paragraph{(iv)}% Suppose $A \cap B = A$. Then every member of $A$ is a member of $B$. This immediately implies $A \subseteq B$. \end{proof} \subsection{\unverified{Exercise 2.18}}% \hyperlabel{sub:exercise-2.18} Assume that $A$ and $B$ are subsets of $S$. List all of the different sets that can be made from these three by use of the binary operations $\cup$, $\cap$, and $-$. \begin{proof} We can reason about this diagrammatically: \begin{figure}[ht] \includegraphics[width=0.6\textwidth]{venn-diagram} \centering \end{figure} In the above diagram, we assume the left circle corresponds to set $A$ and the right circle corresponds to $B$. The the possible sets we can make via the specified operators are: \begin{itemize} \item $A - B$, the left circle excluding the overlapping region. \item $A \cap B$, the overlapping region. \item $B - A$, the right circle excluding the overlapping region. \item $(A \cup B) \cap A$, the left circle. \item $(A \cup B) \cap B$, the right circle. \item $(A - B) \cup (B - A)$, the symmetric difference. \item $A \cup B$, the entire diagram. \end{itemize} \end{proof} \subsection{\verified{Exercise 2.19}}% \hyperlabel{sub:exercise-2.19} Is $\powerset{(A - B)}$ always equal to $\powerset{A} - \powerset{B}$? Is it ever equal to $\powerset{A} - \powerset{B}$? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_19} \begin{proof} Let $A$ and $B$ be arbitrary sets. We show (i) that $\emptyset \in \powerset{(A - B})$ and (ii) $\emptyset \not\in \powerset{A} - \powerset{B}$. \paragraph{(i)}% \hyperlabel{par:exercise-2.19-i} By definition of the \nameref{ref:power-set}, $$\powerset{(A - B)} = \{ x \mid x \subseteq A - B \}.$$ But $\emptyset$ is a subset of \textit{every} set. Thus $\emptyset \in \powerset{(A - B)}$. \paragraph{(ii)}% By the same reasoning found in \nameref{par:exercise-2.19-i}, $\emptyset \in \powerset{A}$ and $\emptyset \in \powerset{B}$. But then, by definition of the relative complement, $\emptyset \not\in \powerset{A} - \powerset{B}$. \paragraph{Conclusion}% By the \nameref{ref:extensionality-axiom}, the two sets are never equal. \end{proof} \subsection{\verified{Exercise 2.20}}% \hyperlabel{sub:exercise-2.20} Let $A$, $B$, and $C$ be sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_20} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. By the \nameref{ref:extensionality-axiom}, $B = C$ if and only if for all sets $x$, $x \in B \iff x \in C$. We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Suppose $x \in B$. Then there are two cases to consider: \subparagraph{Case 1}% Assume $x \in A$. Then $x \in A \cap B$. By hypothesis, $A \cap B = A \cap C$. Thus $x \in A \cap C$ immediately implying $x \in C$. \subparagraph{Case 2}% Assume $x \not\in A$. Then $x \in A \cup B$. By hypothesis, $A \cup B = A \cup C$. Thus $x \in A \cup C$. Since $x \not\in A$, it follows $x \in C$. \paragraph{($\Leftarrow$)}% Suppose $x \in C$. Then there are two cases to consider: \subparagraph{Case 1}% Assume $x \in A$. Then $x \in A \cap C$. By hypothesis, $A \cap B = A \cap C$. Thus $x \in A \cap B$, immediately implying $x \in B$. \subparagraph{Case 2}% Assume $x \not\in A$. Then $x \in A \cup C$. By hypothesis, $A \cup B = A \cup C$. Thus $x \in A \cup B$. Since $x \not\in A$, it follows $x \in B$. \end{proof} \subsection{\verified{Exercise 2.21}}% \hyperlabel{sub:exercise-2.21} Show that $\bigcup\; (A \cup B) = \bigcup A \cup \bigcup B$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_21} \begin{proof} Let $A$ and $B$ be arbitrary sets. By the \nameref{ref:extensionality-axiom}, the specified equality holds if and only if for all sets $x$, $$x \in \bigcup (A \cup B) \iff x \in \bigcup A \cup \bigcup B.$$ We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Suppose $x \in \bigcup (A \cup B)$. By definition of the union of sets, there exists some $b \in A \cup B$ such that $x \in b$. If $b \in A$, then $x \in \bigcup A$ and $x \in \bigcup A \cup \bigcup B$. Alternatively, if $b \in B$, then $x \in \bigcup B$ and $x \in \bigcup A \cup \bigcup B$. Regardless, $x$ is in the target set. \paragraph{($\Leftarrow$)}% Suppose $x \in \bigcup A \cup \bigcup B$. Then $x \in \bigcup A$ or $x \in \bigcup B$. WLOG, suppose $x \in \bigcup A$. By definition of the union of sets, there exists some $b \in A$ such that $x \in b$. But then $b \in A \cup B$ meaning $x$ is also a member of $\bigcup (A \cup B)$. \end{proof} \subsection{\verified{Exercise 2.22}}% \hyperlabel{sub:exercise-2.22} Show that if $A$ and $B$ are nonempty sets, then $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_22} \begin{proof} Let $A$ and $B$ be arbitrary, nonempty sets. By the \nameref{ref:extensionality-axiom}, the specified equality holds if and only if for all sets $x$, \begin{equation} \hyperlabel{sub:exercise-2.22-eq1} x \in \bigcap (A \cup B) \iff x \in \bigcap A \cap \bigcap B. \end{equation} We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Suppose $x \in \bigcap (A \cup B)$. Then for all $b \in A \cup B$, $x \in B$. In other words, for every member $b_1$ of $A$ and every member $b_2$ of $B$, $x$ is a member of both $b_1$ and $b_2$. But that implies $x \in \bigcap A$ and $x \in \bigcap B$. \paragraph{($\Leftarrow$)}% Suppose $x \in \bigcap A \cap \bigcap B$. That is, $x \in \bigcap A$ and $x \in \bigcap B$. By definition of the intersection of sets, forall sets $b$, if $b \in A$, then $x \in b$. Likewise, if $b \in B$, then $x \in b$. In other words, if $b$ is a member of either $A$ or $B$, $x \in b$. That immediately implies $x \in \bigcap (A \cup B$. \end{proof} \subsection{\unverified{Exercise 2.23}}% \hyperlabel{sub:exercise-2.23} Show that if $\mathscr{B}$ is nonempty, then $A \cup \bigcap \mathscr{B} = \bigcap\; \{A \cup X \mid X \in \mathscr{B} \}$. \begin{proof} Refer to \nameref{sub:general-distributive-laws}. \end{proof} \subsection{\verified{Exercise 2.24a}}% \hyperlabel{sub:exercise-2.24a} Show that if $\mathscr{A}$ is nonempty, then $\powerset{\bigcap\mathscr{A}} = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A} \}$. \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_24a} \begin{proof} Suppose $\mathscr{A}$ is a nonempty set. Then $\bigcap \mathscr{A}$ is well-defined. Therefore \begin{align*} \powerset{\bigcap\mathscr{A}} & = \{ x \mid x \subseteq \bigcap \mathscr{A} \} & \textref{ref:power-set} \\ & = \{ x \mid x \subseteq \{ y \mid \forall X \in \mathscr{A}, y \in X \} \} & \text{def'n intersection} \\ & = \{ x \mid \forall t \in x, t \in \{ y \mid \forall X \in \mathscr{A}, y \in X \} \} & \text{def'n subset} \\ & = \{ x \mid \forall t \in x, (\forall X \in \mathscr{A}, t \in X) \} \\ & = \{ x \mid \forall X \in \mathscr{A}, (\forall t \in x, t \in X) \} \\ & = \{ x \mid \forall X \in \mathscr{A}, x \subseteq X \} \\ & = \{ x \mid \forall X \in \mathscr{A}, x \in \powerset{X} \} & \textref{ref:power-set-axiom} \\ & = \{ x \mid \forall t \in \{ \powerset{X} \mid X \in \mathscr{A} \}, x \in t \} \\ & = \bigcap\; \{\powerset{X} \mid X \in \mathscr{A}\}. \end{align*} \end{proof} \subsection{\verified{Exercise 2.24b}}% \hyperlabel{sub:exercise-2.24b} Show that \begin{equation} \hyperlabel{sub:exercise-2.24b-eq1} \bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \} \subseteq \powerset{\bigcup\mathscr{A}}. \end{equation} Under what conditions does equality hold? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_24b} \begin{proof} We first prove \eqref{sub:exercise-2.24b-eq1}. Let $x \in \bigcup\; \{ \powerset{X} \mid X \in \mathscr{A} \}$. By definition of the union of sets, $(\exists X \in \mathscr{A}), x \in \powerset{X}$. By definition of the \nameref{ref:power-set}, $x \subseteq X$. By \nameref{sub:exercise-2.3}, $X \subseteq \bigcup \mathscr{A}$. Therefore $x \subseteq \bigcup \mathscr{A}$, proving $x \in \powerset{\mathscr{A}}$ as expected. \suitdivider \noindent We show $\powerset{\bigcup A} \subseteq \bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$ if and only if $\bigcup\mathscr{A} \in \mathscr{A}$. \paragraph{($\Rightarrow$)}% Suppose $\powerset{\bigcup\mathscr{A}} \subseteq \bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$. By definition of the \nameref{ref:power-set}, $\bigcup\mathscr{A} \in \powerset{\bigcup\mathscr{A}}$. By hypothesis, $\bigcup\mathscr{A} \in \bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$. By definition of the union of sets, there exists some $X \in \mathscr{A}$ such that $\bigcup\mathscr{A} \in \powerset{X}$. That is, $\bigcup\mathscr{A} \subseteq X$. But $\bigcup\mathscr{A}$ cannot be a proper subset of $X$ since $X \in \mathscr{A}$. Thus $\bigcup\mathscr{A} = X$. This proves $\bigcup\mathscr{A} \in \bigcup\;\{ \powerset{X} \mid X \in \mathscr{A} \}$. \paragraph{($\Leftarrow$)}% Suppose $\bigcup\mathscr{A} \in A$. Let $x \in \powerset{\bigcup\mathscr{A}}$. Since $\bigcup\mathscr{A} \in \mathscr{A}$, it immediately follows that $x \in \{\powerset{X} \mid X \in \mathscr{A}\}$. \paragraph{Conclusion}% Equality follows immediately from this fact in conjunction with the proof of \eqref{sub:exercise-2.24b-eq1}. \end{proof} \subsection{\verified{Exercise 2.25}}% \hyperlabel{sub:exercise-2.25} Is $A \cup \bigcup \mathscr{B}$ always the same as $\bigcup\;\{ A \cup X \mid X \in \mathscr{B} \}$? If not, then under what conditions does equality hold? \code*{Bookshelf/Enderton/Set/Chapter\_2} {Enderton.Set.Chapter\_2.exercise\_2\_25} \begin{proof} We prove that \begin{equation} \hyperlabel{sub:exercise-2.25-eq1} A \cup \bigcup \mathscr{B} = \bigcup\;\{ A \cup X \mid X \in \mathscr{B} \} \end{equation} if and only if $A = \emptyset$ or $\mathscr{B} \neq \emptyset$. We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Suppose \eqref{sub:exercise-2.25-eq1} holds true. There are two cases to consider: \subparagraph{Case 1}% Suppose $B \neq \emptyset$. Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially. \subparagraph{Case 2}% Suppose $B = \emptyset$. Then $$A \cup \bigcup \mathscr{B} = A \cup \bigcup \emptyset = A$$ and $$ \bigcup\;\{ A \cup X \mid X \in \mathscr{B} \} = \bigcup \emptyset \\ = \emptyset. $$ Then by hypothesis \eqref{sub:exercise-2.25-eq1}, $A = \emptyset$. Then $A = \emptyset \lor \mathscr{B} \neq \emptyset$ holds trivially. \paragraph{($\Leftarrow$)}% Suppose $A = \emptyset$ or $\mathscr{B} \neq \emptyset$. There are two cases to consider: \paragraph{Case 1}% Suppose $A = \emptyset$. Then $A \cup \bigcup \mathscr{B} = \bigcup{\mathscr{B}}$. Likewise, $$ \bigcup \{ A \cup X \mid X \in \mathscr{B} \} = \bigcup \{ X \mid X \in \mathscr{B} \} \\ = \bigcup \mathscr{B}. $$ Therefore \eqref{sub:exercise-2.25-eq1} holds. \paragraph{Case 2}% Suppose $B \neq \emptyset$. Then \begin{align*} A \cup \bigcup\mathscr{B} & = \{ x \mid x \in A \lor x \in \bigcup\mathscr{B} \} \\ & = \{ x \mid x \in A \lor (\exists b \in \mathscr{B}) x \in b \} \\ & = \{ x \mid (\exists b \in \mathscr{B}) x \in A \lor x \in b \} \\ & = \{ x \mid (\exists b \in \mathscr{B}) x \in A \cup b \} \\ & = \{ x \mid x \in \bigcup \{ A \cup X \mid X \in \mathscr{B} \} \\ & = \bigcup \{ A \cup X \mid X \in \mathscr{B} \}. \end{align*} Therefore \eqref{sub:exercise-2.25-eq1} holds. \end{proof} \chapter{Relations and Functions}% \hyperlabel{chap:relations-functions} \section{Ordered Pairs}% \hyperlabel{sec:ordered-pairs} \subsection{\verified{Theorem 3A}}% \hyperlabel{sub:theorem-3a} \begin{theorem}[3A] For any sets $x$, $y$, $u$, and $v$, \begin{equation} \hyperlabel{sub:theorem-3a-eq1} \tuple{u, v} = \tuple{x, y} \iff u = x \land v = y. \end{equation} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.OrderedPair.ext\_iff} \begin{proof} Let $x$, $y$, $u$, and $v$ be arbitrary sets. \paragraph{($\Leftarrow$)}% This follows trivially. \paragraph{($\Rightarrow$)}% Suppose $\tuple{u, v} = \tuple{x, y}$. Then, by definition of an \nameref{ref:ordered-pair}, \begin{equation} \hyperlabel{sub:theorem-3a-eq2} \{\{u\}, \{u, v\}\} = \{\{x\}, \{x, y\}\}. \end{equation} By the \nameref{ref:extensionality-axiom}, it follows $\{u\} \in \{\{x\}, \{x, y\}\}$ and $\{u, v\} \in \{\{x\}, \{x, y\}\}$. That is, $$\{u\} = \{x\} \quad\text{or}\quad \{u\} = \{x, y\}$$ and $$\{u, v\} = \{x\} \quad\text{or}\quad \{u, v\} = \{x, y\}.$$ There are 4 cases to consider: \paragraph{Case 1}% Suppose $\{u\} = \{x\}$ and $\{u, v\} = \{x\}$. The former identity implies $u = x$. The latter identity implies $u = v = x$. Then \eqref{sub:theorem-3a-eq2} simplifies to $$\{\{u\}\} = \{\{x\}, \{x, y\}\},$$ meaning $x = y$. Thus $v = y$ as well. \paragraph{Case 2}% Suppose $\{u\} = \{x\}$ and $\{u, v\} = \{x, y\}$. The former identity implies $u = x$. Substituting into the latter identity yields $\{u, v\} = \{u, y\}$. This holds if and only if $v = y$. \paragraph{Case 3}% Suppose $\{u\} = \{x, y\}$ and $\{u, v\} = \{x\}$. The former identity implies $x = y = u$. Substituting into the latter yields $\{u, v\} = \{u\}$. Thus $u = v$ which in turn implies $v = y$. \paragraph{Case 4}% Suppose $\{u\} = \{x, y\}$ and $\{u, v\} = \{x, y\}$. The former identity implies $x = y = u$. Substituting into the latter yields $\{u, v\} = \{u\}$. This implies $v = u$ which in turn implies $v = y$. \paragraph{Conclusion}% These cases are exhaustive and each implies that $u = x$ and $v = y$. \end{proof} \subsection{\verified{Lemma 3B}}% \hyperlabel{sub:lemma-3b} \begin{lemma}[3B] If $x \in C$ and $y \in C$, then $\tuple{x, y} \in \powerset{\powerset{C}}$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.lemma\_3b} \begin{proof} Let $C$ be an arbitrary set and $x, y \in C$. Then, by definition of the \nameref{ref:power-set}, $\{x\}$ and $\{x, y\}$ are members of $\powerset{C}$. Likewise, $\{\{x\}, \{x, y\}\}$ is a member of $\powerset{\powerset{C}}$. By definition of an \nameref{ref:ordered-pair}, $\tuple{x, y} = \{\{x\}, \{x, y\}\}$. This concludes our proof. \end{proof} \subsection{\unverified{Corollary 3C}}% \hyperlabel{sub:corollary-3c} \begin{theorem}[3C] For any sets $A$ and $B$, there is a set whose members are exactly the pairs $\tuple{x, y}$ with $x \in A$ and $y \in B$. \end{theorem} \lean{Mathlib/SetTheory/ZFC/Basic}{Set.prod} \begin{proof} Define $C = A \cup B$. Then for all $x \in A$ and for all $y \in B$, $x$ and $y$ are both in $C$. By \nameref{sub:lemma-3b}, it follows that $\tuple{x, y} \in \powerset{\powerset{C}}$. The \nameref{ref:power-set-axiom} indicates $\powerset{\powerset{C}}$ is indeed a set. Therefore the \nameref{ref:subset-axioms} are applicable. This implies the existence of a set $D$ such that $$\forall z, (z \in D \iff z \in \powerset{\powerset{C}} \land (\exists x, \exists y, x \in A \land y \in B \land z = \tuple{x, y})).$$ By construction $D$ is the set whose members are exactly the pairs $\tuple{x, y}$ with $x \in A$ and $y \in B$. \end{proof} \section{Relations}% \hyperlabel{sec:relations} \subsection{\verified{Theorem 3D}}% \hyperlabel{sub:theorem-3d} \begin{theorem}[3D] If $\tuple{x, y} \in A$, then $x$ and $y$ belong to $\bigcup\bigcup A$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3d} \begin{proof} Let $A$ be a set and $\tuple{x, y} \in A$. By definition of an \nameref{ref:ordered-pair}, $$\tuple{x, y} = \{\{x\}, \{x, y\}\}.$$ By \nameref{sub:exercise-2.3}, $\{\{x\}, \{x, y\}\} \subseteq \bigcup A$. Then $\{x, y\} \in \bigcup A$. Another application of \nameref{sub:exercise-2.3} implies $\{x, y\} \subseteq \bigcup\bigcup A$. Therefore $x, y \in \bigcup\bigcup A$. \end{proof} \section{Functions}% \hyperlabel{sec:functions} \subsection{\verified{Theorem 3E}}% \hyperlabel{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} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.dom\_inv\_eq\_ran\_self} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.ran\_inv\_eq\_dom\_self} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.inv\_inv\_eq\_self} \begin{proof} We prove that (i) $\dom{(F^{-1})} = \ran{F}$, (ii) $\ran{(F^{-1})} = \dom{F}$, and (iii) $(F^{-1})^{-1} = F$. \paragraph{(i)}% By definition of the \nameref{ref:domain}, $x \in \dom{(F^{-1})}$ if and only if there exists some $y$ such that $\tuple{x, y} \in F^{-1}$. By definition of the \nameref{ref:inverse} of a set, $\tuple{y, x} \in F$. By definition of the \nameref{ref:range}, $x \in \ran{F}$. Since each step holds biconditionally, it follows $\dom{(F^{-1})} = \ran{F}$ as expected. \paragraph{(ii)}% By definition of the \nameref{ref:range}, $x \in \ran{(F^{-1})}$ if and only if there exists some $t$ such that $\tuple{t, x} \in F^{-1}$. By definition of the \nameref{ref:inverse} of a set, $\tuple{x, t} \in F$. By definition of the \nameref{ref:domain}, $x \in \dom{F}$. Since each step holds biconditionally, it follows $\ran{(F^{-1})} = \dom{F}$. \paragraph{(iii)}% By definition of the \nameref{ref:inverse} of a set, \begin{align*} (F^{-1})^{-1} & = \{\tuple{u, v} \mid \tuple{v, u} \in F^{-1}\} \\ & = \{\tuple{u, v} \mid \tuple{u, v} \in F\} \\ & = F. \end{align*} \end{proof} \subsection{\verified{Theorem 3F}}% \hyperlabel{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} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.single\_valued\_inv\_iff\_single\_rooted\_self} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.single\_valued\_self\_iff\_single\_rooted\_inv} \begin{proof} We prove that (i) any set $F$, $F^{-1}$ is a function iff $F$ is single-rooted and (ii) any relation $F$ is a function iff $F^{-1}$ is single-rooted. \paragraph{(i)}% \hyperlabel{par:theorem-3f-i} Let $F$ be any set. \subparagraph{($\Rightarrow$)}% Suppose $F^{-1}$ is a \nameref{ref:function}. By definition, for each $x \in \dom{(F^{-1})}$, there is only one $y$ such that $\tuple{x, y} \in F^{-1}$. By definition of the \nameref{ref:inverse} of $F$, $F^{-1} = \{\tuple{u, v} \mid vFu\}$. Then for each $x \in \ran{F}$, there exists exactly one $y$ such that $\tuple{y, x} \in F$. This definitionally means $F$ is single-rooted. \subparagraph{($\Leftarrow$)}% Suppose $F$ is single-rooted. By definition, for each $x \in \ran{F}$, there is only one $t$ such that $\tuple{t, x} \in F$. By definition of the \nameref{ref:inverse} of $F$, $F^{-1} = \{\tuple{u, v} \mid vFu\}$. Then for each $x \in \dom{(F^{-1})}$ there exists exactly one $t$ such that $\tuple{x, t} \in F^{-1}$. This definitionally means $F^{-1}$ is a function. \paragraph{(ii)}% Let $F$ be a \nameref{ref:relation}. \subparagraph{($\Rightarrow$)}% Suppose $F$ is a function. By \nameref{sub:theorem-3e}, $F = (F^{-1})^{-1}$. Then by \nameref{par:theorem-3f-i}, $F^{-1}$ is single-rooted. \subparagraph{($\Leftarrow$)}% Suppose $F^{-1}$ is single-rooted. Then by \nameref{par:theorem-3f-i}, $(F^{-1})^{-1}$ is a function. By \nameref{sub:theorem-3e}, $(F^{-1})^{-1} = F$. Thus $F$ is a function. \end{proof} \subsection{\verified{One-to-One Inverse}}% \hyperlabel{sub:one-to-one-inverse} \begin{lemma} For any one-to-one function $F$, $F^{-1}$ is also a one-to-one function. \end{lemma} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.one\_to\_one\_self\_iff\_one\_to\_one\_inv} \begin{proof} We prove that (i) $F^{-1}$ is a function and (ii) $F^{-1}$ is single-rooted. \paragraph{(i)}% \hyperlabel{par:lemma-1-i} By hypothesis, $F$ is one-to-one. This means it is single-rooted, i.e. for all $x \in \ran{F}$, there exists exactly one $t$ such that $\tuple{t, x} \in F$. By definition of the \nameref{ref:inverse} of $F$, $\tuple{x, t} \in F^{-1}$. But then for all $x \in \dom{(F^{-1})}$, there exists exactly one $t$ such that $\tuple{x, t} \in F^{-1}$. Thus $F^{-1}$ is a function. \paragraph{(ii)}% \hyperlabel{par:lemma-1-ii} By hypothesis, $F$ is single-valued. That is, for all $x \in \dom{F}$, there exists exactly one $y$ such that $\tuple{x, y} \in F$. By definition of the \nameref{ref:inverse} of $F$, $\tuple{y, x} \in F^{-1}$. But then for all $x \in \ran{(F^{-1})}$, there exists exactly one $y$ such that $\tuple{y, x} \in F^{-1}$. Thus $F^{-1}$ is single-rooted. \paragraph{Conclusion}% By \nameref{par:lemma-1-i} and \nameref{par:lemma-1-ii}, $F^{-1}$ is a one-to-one function. \end{proof} \subsection{\verified{Theorem 3G}}% \hyperlabel{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} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3g\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3g\_ii} \begin{proof} Suppose $F$ is a one-to-one \nameref{ref:function}. Then \nameref{sub:one-to-one-inverse} indicates $F^{-1}$ is a one-to-one function with domain $\ran{F}$ and range $\dom{F}$. For all $x \in \dom{F}$, $\tuple{x, F(x)} \in F$. Then $\tuple{F(x), x} \in F^{-1}$. Since $F^{-1}$ is single-valued, $F^{-1}(F(x)) = x$. For all $y \in \ran{F}$, $\tuple{y, F^{-1}(y)} \in F^{-1}$. Then $\tuple{F^{-1}(y), y} \in F$. Since $F$ is single-valued, $F(F^{-1}(y)) = y$. \end{proof} \subsection{\verified{Theorem 3H}}% \hyperlabel{sub:theorem-3h} \begin{theorem}[3H] Assume that $F$ and $G$ are functions. Then $F \circ G$ is a function, its domain is \begin{equation} \hyperlabel{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} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.single\_valued\_comp\_is\_single\_valued} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3h\_dom} \begin{proof} Let $F$ and $G$ be \nameref{ref:function}s. By definition of the \nameref{ref:composition} of $F$ and $G$, \begin{equation} \hyperlabel{sub:theorem-3h-eq2} F \circ G = \{\tuple{u, v} \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 $\tuple{x, y} \in F \circ G$. 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)}% \hyperlabel{par:theorem-3h-i} By \eqref{sub:theorem-3h-eq2}, there exists some $t$ such that $\tuple{x, t} \in G$ and $\tuple{t, y} \in F$. 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 $$\tuple{x, G(x)} \in G \quad\text{and}\quad \tuple{G(x), y} \in F.$$ This immediately implies \eqref{sub:theorem-3h-eq1} holds true. \paragraph{(ii)}% Let $x \in \dom{(F \circ G)}$. By definition, $\tuple{x, (F \circ G)(x)} \in F \circ G$. Then \eqref{sub:theorem-3h-eq2} implies $(F \circ G)(x)$ satisfies $\tuple{G(x), (F \circ G)(x)} \in F$. This is equivalent to saying $F(G(x)) = (F \circ G)(x)$ as expected. \end{proof} \subsection{\verified{One-to-One Composition}}% \hyperlabel{sub:one-to-one-composition} \begin{lemma} Let $F$ and $G$ be one-to-one functions. Then $F \circ G$ is also a one-to-one function. \end{lemma} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.one\_to\_one\_comp\_is\_one\_to\_one} \lean{Mathlib/Data/Set/Function} {Set.InjOn.comp} \begin{proof} Let $F \colon B \rightarrow C$ and $G \colon A \rightarrow B$ be one-to-one \nameref{ref:function}s from sets $A$, $B$, and $C$. By definition of the \nameref{ref:composition} of functions, \begin{equation} \hyperlabel{sub:one-to-one-composition-eq1} F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}. \end{equation} By \nameref{sub:theorem-3h}, $F \circ G$ is a function. All that remains is proving $F \circ G$ is one-to-one. Let $(F \circ G)(x_1) = (F \circ G)(x_2) = y$. By \eqref{sub:one-to-one-composition-eq1}, there exists some $t_1$ such that $x_1Gt_1$ and $t_1Fy$. Likewise, there exists some $t_2$ such that $x_2Gt_2$ and $t_2Fy$. Since $F$ is one-to-one, it follows $t_1 = t_2$. Then, since $G$ is also one-to-one, it follows $x_1 = x_2$. Hence $F \circ G$ is one-to-one. \end{proof} \subsection{\verified{Theorem 3I}}% \hyperlabel{sub:theorem-3i} \begin{theorem}[3I] For any sets $F$ and $G$, $$(F \circ G)^{-1} = G^{-1} \circ F^{-1}.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.comp\_inv\_eq\_inv\_comp\_inv} \begin{proof} By definition of the \nameref{ref:composition} of $F$ and $G$, $$F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}.$$ By definition of the \nameref{ref:inverse} of a function, \begin{align*} (F \circ G)^{-1} & = \{\tuple{u, v} \mid \exists t (vGt \land tFu)\} \\ & = \{\tuple{u, v} \mid \exists t (tFu \land vGt)\} \\ & = \{\tuple{u, v} \mid \exists t \left[ u(F^{-1})t \land t(G^{-1})v \right]\} \\ & = G^{-1} \circ F^{-1}. \end{align*} \end{proof} \subsection{\verified{Theorem 3J (a)}}% \hyperlabel{sub:theorem-3j-a} \begin{theorem}[3J(a)] Assume that $F \colon A \rightarrow B$, and that $A$ is nonempty. 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. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3j\_a} \begin{proof} Let $F$ be a \nameref{ref:function} from nonempty set $A$ to set $B$. \subparagraph{($\Rightarrow$)}% Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$. All that remains is to prove $F$ is single-rooted. Let $y \in \ran{F}$. By definition of the \nameref{ref:range} of a function, there exists some $x_1$ such that $\tuple{x_1, y} \in F$. Suppose there exists a set $x_2$ such that $\tuple{x_2, y} \in F$. By hypothesis, $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$. Thus $x_1 = x_2$. Therefore $F$ must be single-rooted. \subparagraph{($\Leftarrow$)}% 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:one-to-one-inverse} 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}. \end{proof} \subsection{\unverified{Theorem 3J (b)}}% \hyperlabel{sub:theorem-3j-b} \begin{theorem}[3J(b)] Assume that $F \colon A \rightarrow B$, and that $A$ is nonempty. 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 $A$ \textit{onto} $B$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3j\_b} \begin{proof} Let $F$ be a \nameref{ref:function} from nonempty set $A$ to set $B$. \subparagraph{($\Rightarrow$)}% 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 $A$ \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 $\tuple{x, y} \in F$. 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 $y \in B$, $\tuple{y, H(y)} \in H \subseteq F^{-1}$ meaning $\tuple{H(y), y} \in F$. Thus $F(H(y)) = y$ as expected. \end{proof} \subsection{\unverified{Bijections and Inverses}}% \hyperlabel{sub:bijections-inverses} \begin{corollary} A function $f$ is a one-to-one correspondence if and only if it has a left and right inverse. \end{corollary} \begin{proof} By definition, a one-to-one correspondence $f$ between sets $A$ and $B$ must be both one-to-one and onto. By \nameref{sub:theorem-3j-a}, $f$ is one-to-one if and only if it has a left inverse. By \nameref{sub:theorem-3j-b}, $f$ is onto $B$ if and only if it has a right inverse. \end{proof} \subsection{\unverified{Left and Right Inverses and Two-Sided Inverses}}% \hyperlabel{sub:left-right-inverse-two-sided-inverse} \begin{lemma} Let $f$ be a function with left inverse $g_1$ and right inverse $g_2$. Then $g_1 = g_2 = f^{-1}$. \end{lemma} \begin{proof} Let $I$ denote the identity map with appropriate domain and codomain depending on placement in the following: \begin{align*} g_1 & = g_1 \circ I \\ & = g_1 \circ (f \circ g_2) \\ & = (g_1 \circ f) \circ g_2 \\ & = I \circ g_2 \\ & = g_2. \end{align*} By \nameref{sub:bijections-inverses}, $f$ is a bijection meaning $f^{-1}$ is both a left and right inverse. Hence $g_1 = g_2 = f^{-1}$. \end{proof} \subsection{\verified{Theorem 3K(a)}}% \hyperlabel{sub:theorem-3k-a} \begin{theorem}[3K(a)] The following hold for any sets. ($F$ need not be a function.) The image of a union is the union of the images: \begin{equation} \hyperlabel{sub:theorem-3k-a-eq1} \img{F}{A \cup B} = \img{F}{A} \cup \img{F}{B} \end{equation} and \begin{equation} \hyperlabel{sub:theorem-3k-a-eq2} \img{F}{\bigcup{\mathscr{A}}} = \bigcup\;\{\img{F}{A} \mid A \in \mathscr{A}\}. \end{equation} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3k\_a} \begin{proof} 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{\verified{Theorem 3K(b)}}% \hyperlabel{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} \hyperlabel{sub:theorem-3k-b-eq1} \img{F}{A \cap B} \subseteq \img{F}{A} \cap \img{F}{B} \end{equation} and \begin{equation} \hyperlabel{sub:theorem-3k-b-eq2} \img{F}{\bigcap\mathscr{A}} \subseteq \bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}. \end{equation} for nonempty $\mathscr{A}$. Equality holds if $F$ is single-rooted. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3k\_b\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3k\_b\_ii} \begin{proof} Let $F$, $A$, $B$ be arbitrary sets. Let $\mathscr{A}$ be a nonempty set. 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)}% \hyperlabel{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)}% \hyperlabel{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 and $\mathscr{A}$ is nonempty, 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{\verified{Theorem 3K(c)}}% \hyperlabel{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} \hyperlabel{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} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3k\_c\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3k\_c\_ii} \begin{proof} We prove that (i) \eqref{sub:theorem-3k-c-eq1} holds and (ii) equality holds if $F$ is single-rooted. \paragraph{(i)}% \hyperlabel{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 $\tuple{u, v} \in F$. Likewise, $\forall w \in B, \tuple{w, v} \not\in F$. 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{\verified{Corollary 3L}}% \hyperlabel{sub:corollary-3l} \begin{theorem}[3L] For any function $G$ and sets $A$, $B$, and $\mathscr{A}$: \begin{align} \img{G^{-1}}{\bigcup{\mathscr{A}}} & = \bigcup\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\}, \hyperlabel{sub:corollary-3l-eq1} \\ \img{G^{-1}}{\bigcap{\mathscr{A}}} & = \bigcap\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\} \text{ for } \mathscr{A} \neq \emptyset, \hyperlabel{sub:corollary-3l-eq2} \\ \img{G^{-1}}{A - B} & = \img{G^{-1}}{A} - \img{G^{-1}}{B}. \hyperlabel{sub:corollary-3l-eq3} \end{align} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.corollary\_3l\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.corollary\_3l\_ii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.corollary\_3l\_iii} \begin{proof} \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} \section{Equivalence Relations}% \hyperlabel{sec:equivalence-relations} \subsection{\verified{Theorem 3M}}% \hyperlabel{sub:theorem-3m} \begin{theorem}[3M] If $R$ is a symmetric and transitive relation, then $R$ is an equivalence relation on $\fld{R}$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3m} \begin{proof} Suppose $R$ is a \nameref{ref:symmetric} and \nameref{ref:transitive} \nameref{ref:relation}. By definition, the \nameref{ref:field} of $R$ is given by $\fld{R} = \dom{R} \cup \ran{R}$. An \nameref{ref:equivalence-relation} on $\fld{R}$ is, by definition, a binary relation \nameref{ref:reflexive} on $\fld{R}$, symmetric, and transitive. All that remains is to show $R$ is reflexive on $\fld{R}$. Let $x \in \fld{R}$. Then $x \in \dom{R}$ or $x \in \ran{R}$. If $x \in \dom{R}$, there exists some $y$ such that $xRy$. Since $R$ is symmetric, it follows $yRx$. Since $R$ is transitive, it follows $xRx$. If instead $x \in \ran{R}$, there exists some $t$ such that $tRx$. Since $R$ is symmetric, it follows $xRt$. Since $R$ is transitive, it follows $xRx$. Thus $R$ is reflexive on $\fld{R}$. \end{proof} \subsection{\verified{Lemma 3N}}% \hyperlabel{sub:lemma-3n} \begin{lemma}[3N] Assume that $R$ is an equivalence relation on $A$ and that $x$ and $y$ belong to $A$. Then $$[x]_R = [y]_R \iff xRy.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.neighborhood\_iff\_mem} \begin{proof} Suppose $R$ is an \nameref{ref:equivalence-relation} on set $A$. Let $x, y \in A$. \paragraph{($\Rightarrow$)}% Suppose $[x]_R = [y]_R$. Since $R$ is an equivalence relation, it is reflexive on $A$. Thus $yRy$ meaning $y \in [y]_R = \{t \mid yRt\}$. Since $[x]_R = [y]_R$, $y \in \{t \mid xRt\}$ as well. That is, $xRy$. \paragraph{($\Leftarrow$)}% Suppose $xRy$. We show $[x]_R \subseteq [y]_R$ and $[y]_R \subseteq [x]_R$. \subparagraph{($\subseteq$)}% Let $t \in [x]_R$. Then $xRt$. Since $R$ is symmetric, $xRy$ implies $yRx$. Since $R$ is transitive, $yRx$ and $xRt$ implies $yRt$. Thus $t \in [y]_R$. \subparagraph{($\supseteq$)}% Let $t \in [y]_R$. Then $yRt$. Since $R$ is transitive, $xRy$ and $yRt$ implies $xRt$. Thus $t \in [x]_R$. \end{proof} \subsection{\verified{Theorem 3P}}% \hyperlabel{sub:theorem-3p} \begin{theorem}[3P] Assume that $R$ is an equivalence relation on $A$. Then the set $\{[x]_R \mid x \in A\}$ of all equivalence classes is a partition of $A$. \end{theorem} \code{Bookshelf/Enderton/Set/Relation} {Set.Relation.modEquiv\_partition} \begin{proof} Let $\Pi = \{[x]_R \mid x \in A\}$. We show that (i) there are no empty sets in $\Pi$, (ii) no two different sets in $\Pi$ have any common elements and (iii) that each element of $A$ is in some set in $\Pi$. \paragraph{(i)}% By construction, every element of $\Pi$ is of form $[x]_R$ for some $x \in A$. At the very least, $x \in A$ is also in $[x]_R$. Thus every element of $\Pi$ must be nonempty. \paragraph{(ii)}% Let $[x]_R, [y]_R \in \Pi$ be two different sets. We must show that $[x]_R \cap [y]_R = \emptyset$. For the sake of contradiction, suppose $[x]_R \cap [y]_R \neq \emptyset$. Let $z \in [x]_R \cap [y]_R$. Then $xRz$ and $yRz$. Since $R$ is an \nameref{ref:equivalence-relation} on $A$, it is \nameref{ref:symmetric} and \nameref{ref:transitive}. Then $zRy$ and $xRy$. By \nameref{sub:lemma-3n}, $xRy$ if and only if $[x]_R = [y]_R$, contradicting the distinctness of $[x]_R$ and $[y]_R$. Thus it follows $[x]_R \cap [y]_R] = \emptyset$. \paragraph{(iii)}% Let $x \in A$. Since $R$ is an \nameref{ref:equivalence-relation} on $A$, it follows $xRx$. Thus $x$ is a member of some set in $\Pi$, namely $[x]_R$. \end{proof} \subsection{\unverified{Theorem 3Q}}% \hyperlabel{sub:theorem-3q} \begin{theorem}[3Q] Assume that $R$ is an equivalence relation on $A$ and that $F \colon A \rightarrow A$. If $F$ is compatible with $R$, then there exists a unique $\hat{F} \colon A / R \rightarrow A / R$ such that \begin{equation} \hyperlabel{sub:theorem-3q-eq1} \hat{F}([x]_R) = [F(x)]_R \quad\text{for all } x \text{ in } A. \end{equation} If $F$ is not compatible with $R$, then no such $\hat{F}$ exists. \end{theorem} \begin{proof} Let $R$ be an \nameref{ref:equivalence-relation} on $A$ and $F \colon A \rightarrow A$. Suppose $F$ is \nameref{ref:compatible} with $R$. Next define \nameref{ref:relation} $\hat{F}$ to be $$\hat{F} = \{\tuple{[x]_R, [F(x)]_R} \mid x \in A\}.$$ By construction $\hat{F}$ has domain $A / R$ and $\ran{\hat{F}} \subseteq A / R$. All that remains is proving (i) $\hat{F}$ is single-valued and (ii) $\hat{F}$ is unique. \paragraph{(i)}% Let $[x_1]_R, [x_2]_R \in \dom{\hat{F}}$ such that $[x_1]_R = [x_2]_R$. By definition of $\hat{F}$, $\tuple{[x_1]_R, [F(x_1)]_R} \in \hat{F}$ and $\tuple{[x_2]_R, [F(x_2)]_R} \in \hat{F}$. By \nameref{sub:lemma-3n}, $[x_1]_R = [x_2]_R$ implies $x_1Rx_2$. Since $F$ is compatible, $F(x_1)RF(x_2)$. Another application of \nameref{sub:lemma-3n} implies that $[F(x_1)]_R = [F(x_2)]_R$. Thus $\hat{F}$ is single-valued, i.e. a function. \paragraph{(ii)}% Suppose there exists another function, say $\hat{G}$, that satisfies \eqref{sub:theorem-3q-eq1}. That is, $$\hat{G}([x]_R) = [F(x)]_R \quad\text{for all } x \text{ in } A.$$ Let $x \in A$. Then $\hat{G}([x]_R) = [F(x)]_R$ and $\hat{F}([x]_R) = [F(x)]_R$. Since this holds for all $x \in A$, $\hat{F}$ and $\hat{G}$ agree on all equivalence classes in $A / R$. Hence, by the \nameref{ref:extensionality-axiom}, $\hat{F} = \hat{G}$. \suitdivider \noindent Suppose $F$ is not compatible with $R$. Then there exists some $x, y \in A$ such that $xRy$ and $\neg F(x)RF(y)$. By \nameref{sub:lemma-3n}, $[x]_R = [y]_R$. For the sake of contradiction, suppose a function $\hat{F}$ exists satisfying \eqref{sub:theorem-3q-eq1}. Then $\hat{F}([x]_R) = \hat{F}([y]_R)$ meaning $[F(x)]_R = [F(y)]_R$. Then \nameref{sub:lemma-3n} implies $F(x)RF(y)$, a contradiction. Therefore our original hypothesis must be incorrect. That is, there is no incompatible function $\hat{F}$ satisfying \eqref{sub:theorem-3q-eq1}. \end{proof} \section{Ordering Relations}% \hyperlabel{sec:ordering-relations} \subsection{\verified{Theorem 3R}}% \hyperlabel{sub:theorem-3r} \begin{theorem}[3R] Let $R$ be a linear ordering on $A$. \begin{enumerate}[(i)] \item There is no $x$ for which $xRx$. \item For distinct $x$ and $y$ in $A$, either $xRy$ or $yRx$ (but not both). \end{enumerate} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.theorem\_3r} \begin{proof} Suppose $R$ is a \nameref{ref:linear-ordering} on $A$. \paragraph{(i)}% Let $x \in A$. By definition, $R$ is \nameref{ref:trichotomous}. Then only one of $xRx$ and $x = x$ can hold. Since $x = x$ obviously holds, it follows $\tuple{x, x} \not\in R$. \paragraph{(ii)}% Let $x, y \in A$ such that $x \neq y$. By definition, $R$ is \nameref{ref:trichotomous}. Thus only one of $$xRy, \quad x = y, \quad yRx$$ hold. By hypothesis $x \neq y$ meaning either $xRy$ or $yRx$ (but not both). \end{proof} \section{Exercises 3}% \hyperlabel{sec:exercises-3} \subsection{\verified{Exercise 3.1}}% \hyperlabel{sub:exercise-3.1} Suppose that we attempted to generalize the Kuratowski definitions of ordered pairs to ordered triples by defining $$\tuple{x, y, z}^* = \{\{x\}, \{x, y\}, \{x, y, z\}\}.$$ Show that this definition is unsuccessful by giving examples of objects $u$, $v$, $w$, $x$, $y$, $z$ with $\tuple{x, y, z}^* = \tuple{u, v, w}^*$ but with either $y \neq v$ or $z \neq w$ (or both). \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_1} \begin{proof} Let $x = 1$, $y = 1$, and $z = 2$. Let $u = 1$, $v = 2$, and $w = 2$. Then \begin{align*} \tuple{x, y, z}^* & = \{\{x\}, \{x, y\}, \{x, y, z\}\} \\ & = \{\{1\}, \{1, 1\}, \{1, 1, 2\}\} \\ & = \{\{1\}, \{1, 2\}\}. \end{align*} Likewise \begin{align*} \tuple{u, v, w}^* & = \{\{u\}, \{u, v\}, \{u, v, w\}\} \\ & = \{\{1\}, \{1, 2\}, \{1, 2, 2\}\} \\ & = \{\{1\}, \{1, 2\}\}. \end{align*} Thus $\tuple{x, y, z}^* = \tuple{u, v, w}^*$ but $y \neq v$. \end{proof} \subsection{\verified{Exercise 3.2a}}% \hyperlabel{sub:exercise-3.2a} Show that $A \times (B \cup C) = (A \times B) \cup (A \times C)$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_2a} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. Then by \nameref{sub:corollary-3c} and the definition of the union of sets, \begin{align*} A \times (B \cup C) & = \{ \tuple{x, y} \mid x \in A \land y \in (B \cup C) \} \\ & = \{ \tuple{x, y} \mid x \in A \land (y \in B \lor y \in C) \} \\ & = \{ \tuple{x, y} \mid (x \in A \land y \in B) \lor (x \in A \land y \in C) \} \\ & = \{ \tuple{x, y} \mid (x \in A \land y \in B) \} \cup \{ \tuple{x, y} \mid (x \in A \land y \in C) \} \\ & = (A \times B) \cup (A \times C). \end{align*} \end{proof} \subsection{\verified{Exercise 3.2b}}% \hyperlabel{sub:exercise-3.2b} Show that if $A \times B = A \times C$ and $A \neq \emptyset$, then $B = C$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_2b} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets such that $A \neq \emptyset$. By \nameref{sub:corollary-3c}, \begin{align} A \times B & = \{ \tuple{x, y} \mid x \in A \land y \in B \} & \hyperlabel{sub:exercise-3.2b-eq1} \\ A \times C & = \{ \tuple{x, y} \mid x \in A \land y \in C \}. & \hyperlabel{sub:exercise-3.2b-eq2} \end{align} There are two cases to consider: \paragraph{Case 1}% Suppose $B \neq \emptyset$. Then $A \times B \neq \emptyset$ and $A \times C \neq \emptyset$. Let $\tuple{x, y} \in A \times B$. By \eqref{sub:exercise-3.2b-eq1}, $x \in A$ and $y \in B$. By the \nameref{ref:extensionality-axiom}, $$\tuple{x, y} \in A \times B \iff \tuple{x, y} \in A \times C.$$ Therefore $\tuple{x, y} \in A \times C$. By \eqref{sub:exercise-3.2b-eq2}, $x \in A$ and $y \in C$. Since membership of $y$ in $B$ and in $C$ holds biconditionally, the \nameref{ref:extensionality-axiom} indicates $B = C$. \paragraph{Case 2}% Suppose $B = \emptyset$. Then there is no $\tuple{x, y}$ such that $x \in A$ and $y \in B$. Thus $A \times B = \emptyset$ and $A \times C = \emptyset$. But then there cannot exist an $\tuple{x, y}$ such that $x \in A$ and $y \in C$ either. Since $A \neq \emptyset$, it must be the case that $C = \emptyset$. Thus $B = C$. \end{proof} \subsection{\verified{Exercise 3.3}}% \hyperlabel{sub:exercise-3.3} Show that $A \times \bigcup \mathscr{B} = \bigcup\;\{ A \times X \mid X \in \mathscr{B} \}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_3} \begin{proof} Let $A$ and $\mathscr{B}$ be arbitrary sets. By \nameref{sub:corollary-3c} and the definition of the union of sets, \begin{align*} A \times \bigcup\mathscr{B} & = \{ \tuple{x, y} \mid x \in A \land y \in \bigcup\mathscr{B} \} \\ & = \{ \tuple{x, y} \mid x \in A \land (\exists b \in \mathscr{B}), y \in b \} \\ & = \{ \tuple{x, y} \mid (\exists b \in \mathscr{B}), x \in A \land y \in b \} \\ & = \bigcup\; \{ A \times X \mid X \in \mathscr{B} \}. \end{align*} \end{proof} \subsection{\unverified{Exercise 3.4}}% \hyperlabel{sub:exercise-3.4} Show that there is no set to which every ordered pair belongs. \begin{proof} For the sake of contradiction, suppose there exists a set $A$ to which every ordered pair belongs. That is, for all sets $x$ and $y$, $\tuple{x, y} = \{\{x\}, \{x, y\}\}$ is a member of $A$. By the \nameref{ref:union-axiom}, it follows that $\bigcup\bigcup A$ is the set to which every set belongs. But \nameref{sub:theorem-2a} shows this is impossible. Thus our original assumption was wrong; there exists no set to which every ordered pair belongs. \end{proof} \subsection{\verified{Exercise 3.5a}}% \hyperlabel{sub:exercise-3.5a} Assume that $A$ and $B$ are given sets, and show that there exists a set $C$ such that for any $y$, \begin{equation} \hyperlabel{sub:exercise-3.5a-eq1} y \in C \iff y = \{x\} \times B \text{ for some } x \text{ in } A. \end{equation} In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_5a} \begin{proof} Let $a \in A$. By the \nameref{ref:pairing-axiom}, $\{a\}$ is a set. By \nameref{sub:corollary-3c}, $\{a\} \times B$ is a set. Again by the \nameref{ref:pairing-axiom}, $\{\{a\} \times B\}$ is a set. Next, by another application of \nameref{sub:corollary-3c}, $A \times B$ is a set. By the \nameref{ref:power-set-axiom}, $\powerset{(A \times B)}$ is a set. Thus, by the \nameref{ref:subset-axioms}, the following is also a set: $$C = \{ y \in \powerset{(A \times B)} \mid \exists a \in A, \forall x, \left[ x \in y \iff \exists b \in B, x = \tuple{a, b} \right] \}.$$ We now show that $C$ satisfies \eqref{sub:exercise-3.5a-eq1}. \paragraph{($\Rightarrow$)}% Suppose $y \in C$. Then there exists some $a \in A$ such that $$\forall x, \left[ x \in y \iff \exists b \in B, x = \tuple{a, b} \right].$$ By the \nameref{ref:extensionality-axiom}, \begin{align*} y & = \{ \tuple{a, b} \mid b \in B \} \\ & = \{ \tuple{x, b} \mid x \in \{a\} \land b \in B \} \\ & = \{ \{a\} \times B \}. \end{align*} \paragraph{($\Leftarrow$)}% Suppose $y = \{a\} \times B$ for some $a \in A$. By \nameref{sub:corollary-3c}, $x \in \{a\} \times B$ if and only if $\exists b \in B$ such that $x = \tuple{a, b}$. But then $x \in y$ if and only if $\exists b \in B$ such that $x = \tuple{a, b}$. This immediately proves $y \in C$. \end{proof} \subsection{\verified{Exercise 3.5b}}% \hyperlabel{sub:exercise-3.5b} With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_5b} \begin{proof} Let $A$ and $B$ be arbitrary sets. We want to show that \begin{equation} \hyperlabel{sub:exercise-3.5b-eq1} A \times B = \bigcup\; \{\{x\} \times B \mid x \in A\}. \end{equation} The left-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of \nameref{sub:corollary-3c}. The right-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of \nameref{sub:exercise-3.5a}. We prove the set on each side is a subset of the other. \paragraph{($\subseteq$)}% Let $c \in A \times B$. Then there exists some $a \in A$ and $b \in B$ such that $c = \tuple{a, b}$. Thus $c \in \{a\} \times B$. We also note $\{a\} \times B \in \{\{x\} \times B \mid x \in A\}$, specifically when $x = a$. Therefore, by the \nameref{ref:union-axiom}, $c \in \bigcup\;\{\{x\} \times B \mid x \in A\}$. \paragraph{($\supseteq$)}% Let $c \in \bigcup\; \{\{x\} \times B \mid x \in A\}$. By the \nameref{ref:union-axiom}, there exists some $b \in \{\{x\} \times B \mid x \in A\}$ such that $c \in b$. Then there exists some $x \in A$ such that $b = \{x\} \times B$. Therefore $c \in \{x\} \times B$. But $x \in A$ meaning $c \in A \times B$ as well. \paragraph{Conclusion}% Since we have shown $A \times B \subseteq \bigcup\; \{\{x\} \times B \mid x \in A\}$ and $A \times B \supseteq \bigcup\; \{\{x\} \times B \mid x \in A\}$, it follows \eqref{sub:exercise-3.5b-eq1} is a true identity. \end{proof} \subsection{\verified{Exercise 3.6}}% \hyperlabel{sub:exercise-3.6} Show that a set $A$ is a relation iff $A \subseteq \dom{A} \times \ran{A}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_6} \begin{proof} 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 $a = \tuple{x, y}$. By the definition of the \nameref{ref:domain} and \nameref{ref:range} of $A$, $x \in \dom{A}$ and $y \in \ran{A}$. Thus $a = \tuple{x, y} \in \dom{A} \times \ran{A}$ 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{\verified{Exercise 3.7}}% \hyperlabel{sub:exercise-3.7} Show that if $R$ is a relation, then $\fld{R} = \bigcup\bigcup R$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_7} \begin{proof} 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)}% \hyperlabel{par:exercise-3.7-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 $\tuple{x, y} = \{\{x\}, \{x, y\}\} \in R$. 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 $\tuple{t, x} = \{\{t\}, \{t, x\}\} \in R$. Then $\{t, x\} \in \bigcup R$ and $x \in \bigcup\bigcup R$. \paragraph{(ii)}% \hyperlabel{par:exercise-3.7-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, $T' = \tuple{x, y} = \{\{x\}, \{x, y\}\}$ for some sets $x$ and $y$. Thus $t = x$ or $t = y$. By \nameref{sub:exercise-3.6}, $t \in \dom{R}$ or $t \in \ran{R}$. In other words, $t \in \fld{R}$. \paragraph{Conclusion}% Since \nameref{par:exercise-3.7-i} and \nameref{par:exercise-3.7-ii} hold, $\fld{R} = \bigcup\bigcup{R}$. \end{proof} \subsection{\verified{Exercise 3.8}}% \hyperlabel{sub:exercise-3.8} Show that for any set $\mathscr{A}$: \begin{align} \dom{\bigcup{\mathscr{A}}} & = \bigcup\;\{ \dom{R} \mid R \in \mathscr{A} \}, & \hyperlabel{sub:exercise-3.8-eq1} \\ \ran{\bigcup{\mathscr{A}}} & = \bigcup\;\{ \ran{R} \mid R \in \mathscr{A} \}. & \hyperlabel{sub:exercise-3.8-eq2} \end{align} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_8\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_8\_ii} \begin{proof} We prove (i) \eqref{sub:exercise-3.8-eq1} and then (ii) \eqref{sub:exercise-3.8-eq2}. \paragraph{(i)}% Let $x \in \dom{\bigcup{\mathscr{A}}}$. By definition of a domain, there exists some $y$ such that $\tuple{x, y} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, $\exists y, \exists R \in \mathscr{A}, \tuple{x, y} \in R$. Equivalently, $\exists R \in \mathscr{A}, \exists y, \tuple{x, y} \in R$. By another application of the definition of a domain, $\exists R \in \mathscr{A}, x \in \dom{R}$. By another application of the definition of the union of sets, $x \in \bigcup\;\{ \dom{R} \mid R \in \mathscr{A} \}$. Since membership of these two sets holds biconditionally, it follows \eqref{sub:exercise-3.8-eq1} holds. \paragraph{(ii)}% Let $x \in \ran{\bigcup{\mathscr{A}}}$. By definition of a range, there exists some $t$ such that $\tuple{t, x} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, $\exists t, \exists R \in \mathscr{A}, \tuple{t, x} \in R$. Equivalently, $\exists R \in \mathscr{A}, \exists t, \tuple{t, x} \in R$. By another application of the definition of a range, $\exists R \in \mathscr{A}, x \in \ran{R}$. By another application of the definition of the union of sets, $x \in \bigcup\;\{ \ran{R} \mid R \in \mathscr{A} \}$. Since membership of these two sets holds biconditionally, it follows \eqref{sub:exercise-3.8-eq2} holds. \end{proof} \subsection{\verified{Exercise 3.9}}% \hyperlabel{sub:exercise-3.9} Discuss the result of replacing the union operation by the intersection operation in the preceding problem. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_9\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_9\_ii} \begin{answer} Replacing the union operation with the intersection problem produces the following relationships \begin{align} \dom{\bigcap{\mathscr{A}}} & \subseteq \bigcap\;\{ \dom{R} \mid R \in \mathscr{A} \}, & \hyperlabel{sub:exercise-3.9-eq1} \\ \ran{\bigcap{\mathscr{A}}} & \subseteq \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}. & \hyperlabel{sub:exercise-3.9-eq2} \end{align} We prove (i) \eqref{sub:exercise-3.9-eq1} and then (ii) \eqref{sub:exercise-3.9-eq2}. \paragraph{(i)}% Let $x \in \dom{\bigcap{\mathscr{A}}}$. By definition of the \nameref{ref:domain} of a set, $\exists y, \tuple{x, y} \in \bigcap{\mathscr{A}}$. By definition of the intersection of sets, $\exists y, \forall R \in \mathscr{A}, \tuple{x, y} \in R$. But this implies that $\forall R \in \mathscr{A}, \exists y, \tuple{x, y} \in R$. By another application of the definition of the \nameref{ref:domain} of a set, $\forall R \in \mathscr{A}, x \in \dom{R}$. By another application of the intersection of sets, $x \in \bigcap\;\{ \dom{R} \mid R \in \mathscr{A} \}$. Thus \eqref{sub:exercise-3.9-eq1} holds. \paragraph{(ii)}% Let $x \in \ran{\bigcap{\mathscr{A}}}$. By definition of the \nameref{ref:range} of a set, $\exists t, \tuple{t, x} \in \bigcap{\mathscr{A}}$. By definition of the intersection of sets, $\exists t, \forall R \in \mathscr{A}, \tuple{t, x} \in R$. But this implies that $\forall R \in \mathscr{A}, \exists t, \tuple{t, x} \in R$. By another application of the definition of the \nameref{ref:range} of a set, $\forall R \in \mathscr{A}, x \in \ran{R}$. By another application of the intersection of sets, $x \in \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}$. Thus \eqref{sub:exercise-3.9-eq2} holds. \end{answer} \subsection{\unverified{Exercise 3.10}}% \hyperlabel{sub:exercise-3.10} Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive integer $m$ less than $4$. \begin{answer} Let $\tuple{x_1, x_2, x_3, x_4}$ denote an arbitrary $4$-tuple. Then \begin{align} \tuple{x_1, x_2, x_3, x_4} & = \tuple{\tuple{x_1, x_2, x_3}, x_4} & \hyperlabel{sub:exercise-7.10-eq1} \\ & = \tuple{\tuple{\tuple{x_1, x_2}, x_3}, x_4} & \hyperlabel{sub:exercise-7.10-eq2} \end{align} Here \eqref{sub:exercise-7.10-eq1} is an equivalent ordered $2$-tuple and \eqref{sub:exercise-7.10-eq2} is an equivalent ordered $3$-tuple. Furthermore, $\tuple{x_1, x_2, x_3, x_4} = \tuple{\tuple{x_1, x_2, x_3, x_4}}$, showing it can be represented as an ordered $1$-tuple as well. \end{answer} \subsection{\unverified{Exercise 3.11}}% \hyperlabel{sub:exercise-3.11} Prove the following version (for functions) of the extensionality principle: Assume that $F$ and $G$ are functions, $\dom{F} = \dom{G}$, and $F(x) = G(x)$ for all $x$ in the common domain. Then $F = G$. \lean*{Init/Core}{funext} \begin{proof} Let $F$ and $G$ be functions such that $\dom{F} = \dom{G}$ and $F(x) = G(x)$ for all $x$ in the common domain. We prove that $\tuple{x, y} \in F$ if and only if $\tuple{x, y} \in G$. But this follows immediately: \begin{align*} \tuple{x, y} \in F & \iff y = F(x) \land \tuple{x, F(x)} \in F \\ & \iff y = G(x) \land \tuple{x, G(x)} \in G \\ & \iff \tuple{x, y} \in G. \end{align*} By the \nameref{ref:extensionality-axiom}, $F = G$. \end{proof} \subsection{\verified{Exercise 3.12}}% \hyperlabel{sub:exercise-3.12} Assume that $f$ and $g$ are functions and show that $$f \subseteq g \iff \dom{f} \subseteq \dom{g} \land (\forall x \in \dom{f}) f(x) = g(x).$$ \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_12} \begin{proof} Let $f$ and $g$ be \nameref{ref:function}s. \paragraph{($\Rightarrow$)}% Suppose $f \subseteq g$. Then for all \nameref{ref:ordered-pair}s $\tuple{x, y}$, $\tuple{x, y} \in f$ implies $\tuple{x, y} \in g$. Thus every $x \in \dom{f}$ must be a member of $\dom{g}$. Likewise, by definition of a function, $f$ and $g$ are single-valued. Thus $f(x) = y$ and $g(x) = y$. Since $x$ is an arbitrary element in the domain of $f$, it follows $(\forall x \in \dom{f}) f(x) = y = g(x)$. \paragraph{($\Leftarrow$)}% Suppose $\dom{f} \subseteq \dom{g}$ and $(\forall x \in \dom{f}) f(x) = g(x)$. Let $\tuple{x, y} \in f$. By hypothesis, $x \in \dom{g}$ and $y = f(x) = g(x)$. Thus $\tuple{x, y} \in g$ as well. Therefore $f \subseteq g$. \end{proof} \subsection{\verified{Exercise 3.13}}% \hyperlabel{sub:exercise-3.13} Assume that $f$ and $g$ are functions with $f \subseteq g$ and $\dom{g} \subseteq \dom{f}$. Show that $f = g$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_13} \begin{proof} Let $f$ and $g$ be functions such that $f \subseteq g$ and $\dom{g} \subseteq \dom{f}$. By \nameref{sub:exercise-3.12}, it follows that $\dom{f} \subseteq \dom{g}$ and $(\forall x \in \dom{f}) f(x) = g(x)$. Since $\dom{g} \subseteq \dom{f}$ and $\dom{f} \subseteq \dom{g}$, it follows that $\dom{g} = \dom{f}$. By \nameref{sub:exercise-3.11}, $f = g$. \end{proof} \subsection{\verified{Exercise 3.14}}% \hyperlabel{sub:exercise-3.14} Assume that $f$ and $g$ are functions. \begin{enumerate}[(a)] \item Show that $f \cap g$ is a function. \item Show that $f \cup g$ is a function iff $f(x) = g(x)$ for every $x$ in $(\dom{f}) \cap (\dom{g})$. \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_14\_a} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_14\_b} \begin{proof} Assume $f$ and $g$ are \nameref{ref:function}s. \paragraph{(a)}% Consider $f \cap g$. By definition of the intersection of sets, $f \cap g \subseteq f$. Since $f$ is single-valued, it trivially follows that so must $f \cap g$. Therefore $f \cap g$ is a function. \paragraph{(b)}% \subparagraph{($\Rightarrow$)}% Suppose $f \cup g$ is a function. Let $x \in (\dom{f}) \cap (\dom{g})$. That is, $x \in \dom{f}$ and $x \in \dom{g}$. Then there exists only one $y_1$ such that $\tuple{x, y_1} \in f$. Likewise there exists only one $y_2$ such that $\tuple{x, y_2} \in g$. But $\tuple{x, y_1} \in f \cup g$ and $\tuple{x, y_2} \in f \cup g$. Since $f \cup g$ is single-valued, it follows $y_1 = y_2$. That is, $f(x) = g(x)$. \subparagraph{($\Leftarrow$)}% Suppose $f(x) = g(x)$ for every $x \in (\dom{f}) \cap (\dom{g})$. Let $x \in \dom{(f \cup g)}$. There are three cases to consider: \begin{enumerate}[(i)] \item Suppose $x \in \dom{f}$ but not in $\dom{g}$. Since $f$ is a function, it follows $f \cup g$ has only one value $y$ such that $\tuple{x, y} \in f \cup g$. \item Suppose $x \in \dom{g}$ but not in $\dom{f}$. Again, since $g$ is a function, it follows $f \cup g$ has only one value $y$ such that $\tuple{x, y} \in f \cup g$. \item Suppose $x \in \dom{f}$ and $x \in \dom{g}$. By hypothesis, $f(x) = g(x)$ meaning there is only one value $y$ such that $\tuple{x, y} \in f \cup g$. \end{enumerate} The above cases are exhaustive. Together they imply that $f \cup g$ is single-valued, i.e. a function. \end{proof} \subsection{\verified{Exercise 3.15}}% \hyperlabel{sub:exercise-3.15} Let $\mathscr{A}$ be a set of functions such that for any $f$ and $g$ in $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$. Show that $\bigcup{\mathscr{A}}$ is a function. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_15} \begin{proof} Let $\mathscr{A}$ be a set of \nameref{ref:function}s such that for any $f$ and $g$ in $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$. Let $x \in \dom{\bigcup{\mathscr{A}}}$. Then there exists some $y_1$ such that $\tuple{x, y_1} \in \bigcup{\mathscr{A}}$. Suppose there also exists some $y_2$ such that $\tuple{x, y_2} \in \bigcup{\mathscr{A}}$. By definition of the union of sets, there exists some function $f \in \mathscr{A}$ such that $\tuple{x, y_1} \in f$. Likewise there exists some function $g \in \mathscr{A}$ such that $\tuple{x, y_2} \in g$. There are two cases to consider: \paragraph{Case 1}% Suppose $f \subseteq g$. Then $\tuple{x, y_1}, \tuple{x, y_2} \in g$. Since $g$ is a function, i.e. single-valued, $y_1 = y_2$. \paragraph{Case 2}% Suppose $g \subseteq f$. Then $\tuple{x, y_1}, \tuple{x, y_2} \in f$. Since $f$ is a function, i.e. single-valued, $y_1 = y_2$. \paragraph{Conclusion}% Since the above two cases applies for all $x \in \dom{\bigcup{\mathscr{A}}}$ and appropriate choices of $f$ and $g$, it follows $\bigcup{\mathscr{A}}$ is indeed a function. \end{proof} \subsection{\unverified{Exercise 3.16}}% \hyperlabel{sub:exercise-3.16} Show that there is no set to which every function belongs. \begin{proof} Every \nameref{ref:relation} consisting of a single \nameref{ref:ordered-pair} is, by definition, a \nameref{ref:function}. By \nameref{sub:exercise-3.4}, there is no set to which every ordered pair belongs. Thus there is no set to which every function of the described type belongs either, let alone a set to which \textit{every} function belongs. \end{proof} \subsection{\verified{Exercise 3.17}}% \hyperlabel{sub:exercise-3.17} Show that the composition of two single-rooted sets is again single-rooted. Conclude that the composition of two one-to-one functions is again one-to-one. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_17\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_17\_ii} \begin{proof} Let $F$ and $G$ be two single-rooted sets. Consider $F \circ G$. By definition of the \nameref{ref:composition} of sets, \begin{equation} \hyperlabel{sub:exercise-3.17-eq1} F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}. \end{equation} Consider any $v \in \ran{(F \circ G)}$. By definition of the \nameref{ref:range} of a \nameref{ref:relation}, there exists some $u_1$ such that $\tuple{u_1, v} \in F \circ G$. Let $u_2$ be a set such that $\tuple{u_2, v} \in F \circ G$. By \eqref{sub:exercise-3.17-eq1}, there exists a set $t_1$ such that $\tuple{u_1, t_1} \in G$ and $\tuple{t_1, v} \in F$. Likewise, there exists a set $t_2$ such that $\tuple{u_2, t_2} \in G$ and $\tuple{t_2, v} \in F$. But $F$ is single-rooted, meaning $t_1 = t_2$. Likewise, because $G$ is single-rooted, $u_1 = u_2$. Thus $F \circ G$ must also be single-rooted. \suitdivider Let $f$ and $g$ be one-to-one functions. By \nameref{sub:theorem-3h}, $f \circ g$ is single-valued. By the above, $f \circ g$ is single-rooted. Thus $f \circ g$ is one-to-one. \end{proof} \subsection{\verified{Exercise 3.18}}% \hyperlabel{sub:exercise-3.18} Let $R$ be the set $$\{ \tuple{0, 1}, \tuple{0, 2}, \tuple{0, 3}, \tuple{1, 2}, \tuple{1, 3}, \tuple{2, 3}\}.$$ Evaluate the following: $R \circ R$, $R \restriction \{1\}$, $R^{-1} \restriction \{1\}$, $\img{R}{\{1\}}$, and $\img{R^{-1}}{\{1\}}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_18\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_18\_ii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_18\_iii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_18\_iv} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_18\_v} \begin{proof} \begin{enumerate}[(i)] \item $R \circ R = \{ \tuple{0, 2}, \tuple{0, 3}, \tuple{1, 3} \}$. \item $R \restriction \{1\} = \{ \tuple{1, 2}, \tuple{1, 3} \}$. \item $R^{-1} \restriction \{1\} = \{\tuple{1, 0}\}$. \item $\img{R}{\{1\}} = \{2, 3\}$. \item $\img{R^{-1}}{\{1\}} = \{0\}$. \end{enumerate} \end{proof} \subsection{\verified{Exercise 3.19}}% \hyperlabel{sub:exercise-3.19} Let $$A = \{ \tuple{\emptyset, \{\emptyset, \{\emptyset\}\}}, \tuple{\{\emptyset\}, \emptyset} \}.$$ Evaluate each of the following: $A(\emptyset)$, $\img{A}{\emptyset}$, $\img{A}{\{\emptyset\}}$, $\img{A}{\{\emptyset, \{\emptyset\}\}}$, $A^{-1}$, $A \circ A$, $A \restriction \emptyset$, $A \restriction \{\emptyset\}$, $A \restriction \{\emptyset, \{\emptyset\}\}$, $\bigcup\bigcup A$. \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_ii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_iii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_iv} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_v} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_vi} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_vii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_viii} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_ix} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_19\_x} \begin{proof} \begin{enumerate}[(i)] \item $A(\emptyset) = \{\emptyset, \{\emptyset\}\}$. \item $\img{A}{\emptyset} = \emptyset$. \item $\img{A}{\{\emptyset\}} = \{\{\emptyset, \{\emptyset\}\}\}$. \item $\img{A}{\{\emptyset, \{\emptyset\}\}} = \{\{\emptyset, \{\emptyset\}\}, \emptyset\}$. \item $A^{-1} = \{ \tuple{\{\emptyset, \{\emptyset\}\}, \emptyset}, \tuple{\emptyset, \{\emptyset\}} \}$. \item $A \circ A = \{\tuple{\{\emptyset\}, \{\emptyset, \{\emptyset\}\}}\}$. \item $A \restriction \emptyset = \emptyset$ \item $A \restriction \{\emptyset\} = \{\tuple{\emptyset, \{\emptyset, \{\emptyset\}\}}\}$. \item $A \restriction \{\emptyset, \{\emptyset\}\} = A$. \item $\bigcup\bigcup A = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$. \end{enumerate} \end{proof} \subsection{\verified{Exercise 3.20}}% \hyperlabel{sub:exercise-3.20} Show that $F \restriction A = F \cap (A \times \ran{F})$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_20} \begin{proof} Let $F$ and $A$ be arbitrary sets. By \nameref{sub:corollary-3c} and definition of the \nameref{ref:restriction}, intersection, and \nameref{ref:range} of sets, \begin{align*} F \restriction A & = \{\tuple{u, v} \mid uFv \land u \in A\} \\ & = \{\tuple{u, v} \mid uFv \land u \in A \land v \in \ran{F}\} \\ & = \{\tuple{u, v} \mid uFv\} \cap \{\tuple{u, v} \mid u \in A \land v \in \ran{F}\} \\ & = F \cap \{\tuple{u, v} \mid u \in A \land v \in \ran{F}\} \\ & = F \cap (A \times \ran{F}). \end{align*} \end{proof} \subsection{\verified{Exercise 3.21}}% \hyperlabel{sub:exercise-3.21} Show that $(R \circ S) \circ T = R \circ (S \circ T)$. \code*{Bookshelf/Enderton/Set/Relation} {Set.Relation.comp\_assoc} \begin{proof} Let $R$, $S$, and $T$ be arbitrary sets. By definition of the \nameref{ref:composition} of sets, \begin{align*} (R \circ S) \circ T & = \{\tuple{u, v} \mid \exists t(uTt \land t(R \circ S)v)\} \\ & = \{\tuple{u, v} \mid \exists t(uTt \land (\exists a(tSa \land aRv))\} \\ & = \{\tuple{u, v} \mid \exists t, \exists a, (uTt \land tSa) \land aRv)\} \\ & = \{\tuple{u, v} \mid \exists a, \exists t, (uTt \land tSa) \land aRv)\} \\ & = \{\tuple{u, v} \mid \exists a, (\exists t(uTt \land tSa)) \land aRv)\} \\ & = \{\tuple{u, v} \mid \exists a, u(S \circ T)a \land aRv)\} \\ & = R \circ (S \circ T). \end{align*} \end{proof} \subsection{\verified{Exercise 3.22}}% \hyperlabel{sub:exercise-3.22} Show that the following are correct for any sets. \begin{enumerate}[(a)] \item $A \subseteq B \Rightarrow \img{F}{A} \subseteq \img{F}{B}$. \item $\img{(F \circ G)}{A} = \img{F}{\img{G}{A}}$. \item $Q \restriction (A \cup B) = (Q \restriction A) \cup (Q \restriction B)$. \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_22\_a} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_22\_b} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_22\_c} \begin{proof} Let $A$, $B$, $F$, $G$, and $Q$ be arbitrary sets. \paragraph{(a)}% Suppose $A \subseteq B$. Let $x \in \img{F}{A}$. By definition of the \nameref{ref:image} of a set, $\img{F}{A} = \{v \mid (\exists u \in A) uFv\}$. Thus there exists some $u \in A$ such that $uFx$. But $A \subseteq B$ meaning $u \in B$. That is, $(\exists u \in B)uFx$. Thus $$x \in \{v \mid (\exists u \in B)uFv\} = \img{F}{B}.$$ \paragraph{(b)}% By definition of the \nameref{ref:composition} and \nameref{ref:image} of a set, \begin{align*} \img{(F \circ G)}{A} & = \{v \mid (\exists u \in A) u(F \circ G)v\} \\ & = \{v \mid (\exists u \in A) \tuple{u, v} \in F \circ G\} \\ & = \{v \mid (\exists u \in A) \tuple{u, v} \in \{\tuple{b, c} \mid \exists a(bGa \land aFc)\}\} \\ & = \{v \mid \exists u \in A, \exists a, uGa \land aFv\} \\ & = \{v \mid \exists a, \exists u \in A, uGa \land aFv\} \\ & = \{v \mid \exists a, (\exists u \in A, uGa) \land aFv\} \\ & = \{v \mid \exists a \in \{w \mid (\exists u \in A)uGw\}, aFv\} \\ & = \{v \mid (\exists a \in \img{G}{A}) aFv\} \\ & = \img{F}{\img{G}{A}}. \end{align*} \paragraph{(c)}% By definition of the \nameref{ref:restriction} of a set, \begin{align*} Q \restriction (A \cup B) & = \{\tuple{u, v} \mid uQv \land u \in A \cup B\} \\ & = \{\tuple{u, v} \mid uQv \land (u \in A \lor u \in B)\} \\ & = \{\tuple{u, v} \mid (uQv \land u \in A) \lor (uQv \land u \in B)\} \\ & = \{\tuple{u, v} \mid uQv \land u \in A\} \cup \{\tuple{u, v} \mid uQv \land u \in B\} \\ & = (Q \restriction A) \cup (Q \restriction B). \end{align*} \end{proof} \subsection{\verified{Exercise 3.23}}% \hyperlabel{sub:exercise-3.23} Let $I_A$ be the identity function on the set $A$. Show that for any sets $B$ and $C$, $$B \circ I_A = B \restriction A \quad\text{and}\quad \img{I_A}{C} = A \cap C.$$ \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_23\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_23\_ii} \begin{proof} Let $I_A$ be the identity function on the set $A$. That is, $I_A = \{\tuple{u, u} \mid u \in A\}$. Let $B$ and $C$ be any sets. We show that (i) $B \circ I_A = B \restriction A$ and (ii) $\img{I_A}{C} = A \cap C$. \paragraph{(i)}% We show that $B \circ I_A \subseteq B \restriction A$ and $B \restriction A \subseteq B \circ I_A$. \subparagraph{($\subseteq$)}% Let $\tuple{x, y} \in B \circ I_A$. By definition of the \nameref{ref:composition} of sets, there exists some $t$ such that $x(I_A)t$ and $tBy$. By definition of the identity function, $I_A(x) = t$ implies $x = t$. Thus $xBy$. By hypothesis, $x \in \dom{(B \circ I_A)}$. Therefore $x \in \dom{I_A} = A$. Thus $$\tuple{x, y} \in \{\tuple{u, v} \mid u \in A \land uBv\} = B \restriction A.$$ \subparagraph{($\supseteq$)}% Let $\tuple{x, y} \in B \restriction A$. By definition of the \nameref{ref:restriction} of sets, $x \in A$ and $xBy$. But $I_A(x) = x$ meaning $\tuple{I_A(x), y} \in B$. In other words, $\tuple{x, y} \in B \circ I_A$. \paragraph{(ii)}% By definition of the \nameref{ref:image} of sets, \begin{align*} \img{I_A}{C} & = \{v \mid (\exists u \in C) \tuple{u, v} \in I_A\} \\ & = \{v \mid \exists u \in C, u \in A \land u = v\} \\ & = \{v \mid v \in C \land v \in A\} \\ & = C \cap A. \end{align*} \end{proof} \subsection{\verified{Exercise 3.24}}% \hyperlabel{sub:exercise-3.24} Show that for a function $F$, $\img{F^{-1}}{A} = \{x \in \dom{F} \mid F(x) \in A\}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_24} \begin{proof} Let $F$ be a function. By definition of the \nameref{ref:inverse} of a set, \begin{align*} \img{F^{-1}}{A} & = \{x \mid (\exists y \in A) yF^{-1}x\} \\ & = \{x \mid (\exists y \in A) xFy\} \\ & = \{x \mid (\exists y \in A) \tuple{x, y} \in F\} \\ & = \{x \mid x \in \dom{F} \land F(x) \in A\} \\ & = \{x \in \dom{F} \mid F(x) \in A\}. \end{align*} \end{proof} \subsection{\verified{Exercise 3.25}}% \hyperlabel{sub:exercise-3.25} \begin{enumerate}[(a)] \item Assume that $G$ is a one-to-one function. Show that $G \circ G^{-1}$ is $I_{\ran{G}}$, the identity function on $\ran{G}$. \item Show that the result of part (a) holds for any function $G$, not necessarily one-to-one. \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_25\_b} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_25\_a} \begin{proof} \paragraph{(b)}% \hyperlabel{par:exercise-3.25-b} Let $G$ be an arbitrary function. We show that $G \circ G^{-1} \subseteq I_{\ran{G}}$ and that $I_{\ran{G}} \subseteq G \circ G^{-1}$. \subparagraph{($\subseteq$)}% Let $\tuple{x, y} \in G \circ G^{-1}$. By definition of the \nameref{ref:composition} of sets, there exists some set $t$ such that $x(G^{-1})t$ and $tGy$. By definition of the \nameref{ref:inverse} of a set, $$x(G^{-1})t \iff tGx.$$ The right hand side of the above biconditional indicates $x \in \ran{G}$. Since $G$ is single-valued, $tGy \land tGx$ implies $x = y$. Thus $\tuple{x, y} \in I_{\ran{G}}$. \subparagraph{($\supseteq$)}% Let $\tuple{x, x} \in I_{\ran{G}}$ where $x \in \ran{G}$. By definition of the \nameref{ref:range} of a function, there exists some $t$ such that $\tuple{t, x} \in G$. By definition of the \nameref{ref:inverse} of a set, it follows $\tuple{x, t} \in G^{-1}$. Thus $\tuple{x, x} \in G \circ G^{-1}$. \subparagraph{Conclusion}% Since $G \circ G^{-1}$ is a subset of $I_{\ran{G}}$ and vice versa, it follows that these two sets are equal. \paragraph{(a)}% This immediately follows from part \nameref{par:exercise-3.25-b}. \end{proof} \subsection{\verified{Exercise 3.26}}% \hyperlabel{sub:exercise-3.26} Prove the second halves of parts (a) and (b) of Theorem 3K. \begin{proof} Refer to \nameref{sub:theorem-3k-a}, \nameref{sub:theorem-3k-b}, and \nameref{sub:theorem-3k-c}. \end{proof} \subsection{\verified{Exercise 3.27}}% \hyperlabel{sub:exercise-3.27} Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$. ($F$ and $G$ need not be functions.) \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_27} \begin{proof} Let $F$ and $G$ be arbitrary sets. We show that each side of our desired equality is a subset of the other. \paragraph{($\subseteq$)}% Let $x \in \dom{(F \circ G)}$. Then there exists a set $y$ such that $\tuple{x, y} \in F \circ G$. By definition of the \nameref{ref:composition} of sets, there exists a set $t$ such that $xGt$ and $tFy$. Thus $t \in \dom{F}$. Therefore \begin{align*} x & \in \{v \mid (\exists t \in \dom{F}) vGt\} \\ & = \{v \mid (\exists t \in \dom{F}) t(G^{-1})v\} \\ & = \img{G^{-1}}{\dom{F}}. \end{align*} \paragraph{($\supseteq$)}% Let $x \in \img{G^{-1}}{\dom{F}}$. Then, by definition of the \nameref{ref:image} of a set, there exists some $u \in \dom{F}$ such that $u(G^{-1})x$. By definition of the \nameref{ref:inverse} of a set, $xGu$. By definition of the \nameref{ref:domain} of a set, there exists some $t$ such that $uFt$. Thus $xGu \land uFt$. By definition of the \nameref{ref:composition} of sets, $\tuple{x, t} \in F \circ G$. Therefore $x \in \dom{(F \circ G)}$. \end{proof} \subsection{\verified{Exercise 3.28}}% \hyperlabel{sub:exercise-3.28} Assume that $f$ is a one-to-one function from $A$ into $B$, and that $G$ is the function with $\dom{G} = \powerset{A}$ defined by the equation $G(X) = \img{f}{X}$. Show that $G$ maps $\powerset{A}$ one-to-one into $\powerset{B}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_28} \begin{proof} By construction, $\dom{G} = \powerset{A}$. Likewise, $\ran{G} \subseteq \powerset{B}$ by definition of the \nameref{ref:image} of sets. Thus $G$ maps $\powerset{A}$ into $\powerset{B}$. Let $y \in \ran{G}$. Then there exists an $X_1 \in \powerset{A}$ such that $\img{f}{X_1} = y$. To prove $G$ is one-to-one into $\powerset{B}$, assume there exists an $X_2 \in \powerset{A}$ such that $\img{f}{X_2} = y$. All that remains is showing $X_1 = X_2$. Let $t \in X_1$. By definition of the \nameref{ref:image} of a set, $f(t) \in \img{f}{X_1}$. Since $\img{f}{X_1} = \img{f}{X_2}$, it follows $f(t) \in \img{f}{X_2}$. Because $f$ is one-to-one, $f(t) \in \img{f}{X_2}$ if and only if $t \in X_2$. Thus $t \in X_1$ if and only if $t \in X_2$. By the \nameref{ref:extensionality-axiom}, it follows $X_1 = X_2$. \end{proof} \subsection{\verified{Exercise 3.29}}% \hyperlabel{sub:exercise-3.29} Assume that $f \colon A \rightarrow B$ and define a function $G \colon B \rightarrow \powerset{A}$ by \begin{equation} \hyperlabel{sub:exercise-3.29-eq1} G(b) = \{x \in A \mid f(x) = b\}. \end{equation} Show that if $f$ maps $A$ \textit{onto} $B$, then $G$ is one-to-one. Does the converse hold? \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_29} \begin{proof} Let $f \colon A \rightarrow B$ such that $f$ maps $A$ onto $B$. Define $G \colon B \rightarrow \powerset{A}$ by \eqref{sub:exercise-3.29-eq1}. Let $y \in \ran{G}$. By definition of the \nameref{ref:range} of a set, there exists an $x_1 \in B$ such that $G(x_1) = y$. To prove $G$ is one-to-one, suppose there exists an $x_2 \in B$ such that $G(x_2) = y$. All that remains is proving $x_1 = x_2$. By \eqref{sub:exercise-3.29-eq1}, it follows \begin{align*} G(x_1) & = \{x \in A \mid f(x) = x_1\} \\ G(x_2) & = \{x \in A \mid f(x) = x_2\}. \end{align*} Since $f$ maps $A$ onto $B$, $\ran{f} = B$. Thus $x_1, x_2 \in \ran{f}$. By definition of the \nameref{ref:range} of a set, there exist some $t \in A$ such that $f(t) = x_1$. Therefore $t \in G(x_1)$. By the \nameref{ref:extensionality-axiom}, $t \in G(x_2)$. Then $f(t) = x_2$. But $f$ is a \nameref{ref:function}, i.e. single-valued. Thus $x_1 = x_2$. \suitdivider If $G$ is one-to-one, it does not follow that $f$ maps $A$ onto $B$. As a counterexample, let $f \colon \{1\} \rightarrow \{1, 2\}$ given by $f(x) = x$. Define $G \colon \{1, 2\} \rightarrow \powerset{\{1\}}$ by $$G(b) = \{x \in \{1\} \mid f(x) = b\}.$$ $G$ is trivially one-to-one since $G(1) = \{1\}$ and $G(2) = \emptyset$. But $f$ does not map onto $\{1, 2\}$; there is no element in its domain that corresponds to value $2$. \end{proof} \subsection{\verified{Exercise 3.30}}% \hyperlabel{sub:exercise-3.30} Assume that $F \colon \powerset{A} \rightarrow \powerset{A}$ and that $F$ has the monotonicity property: $$X \subseteq Y \subseteq A \Rightarrow F(X) \subseteq F(Y).$$ Define $$B = \bigcap\{X \subseteq A \mid F(X) \subseteq X\} \quad\text{and}\quad C = \bigcup\{X \subseteq A \mid X \subseteq F(X)\}.$$ \subsubsection{\verified{Exercise 3.30 (a)}}% \hyperlabel{ssub:exercise-3.30-a} Show that $F(B) = B$ and $F(C) = C$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_30\_a} \begin{proof} We prove that (i) $F(B) = B$ and (ii) $F(C) = C$. \paragraph{(i)}% \hyperlabel{par:exercise-3.30-a-i} To prove $F(B) = B$, we show $F(B) \subseteq B$ and $F(B) \supseteq B$. \subparagraph{($\subseteq$)}% \hyperlabel{spar:exercise-3.30-i-sub} Let $x \in F(B)$ and $X \subseteq A$ such that $F(X) \subseteq X$. By definition of $B$, $B \subseteq X$. Then monotonicity implies $F(B) \subseteq F(X) \subseteq X$. Therefore $x \in X$. Since $X$ was arbitrarily chosen, it follows that $$\forall X, X \subseteq A \land F(X) \subseteq X \Rightarrow x \in X.$$ By definition of the intersection of sets, $$x \in \bigcap\{X \subseteq A \mid F(X) \subseteq X\} = B.$$ Hence $F(B) \subseteq B$. \subparagraph{($\supseteq$)}% By \nameref{spar:exercise-3.30-i-sub}, $F(B) \subseteq B$. Then monotonicity implies $F(F(B)) \subseteq F(B)$. Thus $F(B) \in \{X \subseteq A \mid F(X) \subseteq X\}$. Hence, by definition of $B$, $B \subseteq F(B)$. \paragraph{(ii)}% To prove $F(C) = C$, we show $F(C) \supseteq C$ and $F(C) \subseteq C$. \subparagraph{($\supseteq$)}% \hyperlabel{spar:exercise-3.30-i-sup} Let $x \in C$. By definition of $C$, there exists some $X \subseteq A$ such that $X \subseteq F(X)$ and $x \in X$. Since $C$ contains $X$, $X \subseteq C$. Then monotonicity implies $X \subseteq F(X) \subseteq F(C)$. Therefore $x \in F(C)$. \subparagraph{($\subseteq$)}% By \nameref{spar:exercise-3.30-i-sup}, $C \subseteq F(C)$. Then monotonicity implies $F(C) \subseteq F(F(C))$. Thus $F(C) \in \{X \subseteq A \mid X \subseteq F(X)\}$. Hence, by definition of $C$, $F(C) \subseteq C$. \end{proof} \subsubsection{\verified{Exercise 3.30 (b)}}% \hyperlabel{ssub:exercise-3.30-b} Show that if $F(X) = X$, then $B \subseteq X \subseteq C$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_30\_b} \begin{proof} Suppose $F(X) = X$ for some $X \subseteq A$. By the \nameref{ref:extensionality-axiom}, $F(X) \subseteq X$ and $X \subseteq F(X)$. Therefore \begin{align} X & \in \{X \subseteq A \mid F(X) \subseteq X\} & \hyperlabel{ssub:exercise-3.30-b-eq1} \\ X & \in \{X \subseteq A \mid X \subseteq F(X)\}. & \hyperlabel{ssub:exercise-3.30-b-eq2} \end{align} \eqref{ssub:exercise-3.30-b-eq1} immediately implies $B \subseteq X$ and \eqref{ssub:exercise-3.30-b-eq2} immediately implies $X \subseteq C$. Thus $B \subseteq X \subseteq C$. \end{proof} \subsection{\unverified{Exercise 3.31}}% \hyperlabel{sub:exercise-3.31} Show that from the first form of the axiom of choice we can prove the second form, and conversely. \begin{proof} We prove the first form holds if and only if the second form holds. \paragraph{($\Rightarrow$)}% We assume the first form of the axiom of choice. Let $I$ be a set and $H$ be a function with $\dom{H} = I$. Furthermore, suppose $H(i) \neq \emptyset$ for all $i \in I$. By definition of the \nameref{ref:cartesian-product}, $$\bigtimes_{i \in I} H(i) = \{f \mid f \text{ is a function with } \dom{f} = I \text{ and } (\forall i \in I) f(i) \in H(i)\}.$$ Consider the relation $R$ formed by $$R = \bigcup_{i \in I} \{i\} \times H(i).$$ By the \nameref{ref:axiom-of-choice-1}, there exists a function $f \subseteq R$ with $\dom{f} = I$. Furthermore, for all $i \in I$, it must be $f(i) \in H(i)$ by construction. Then $f$ is a member of $\bigtimes_{i \in I} H(i)$. That is, $\bigtimes_{i \in I} H(i) \neq \emptyset$. \paragraph{($\Leftarrow$)}% We assume the second form of the axiom of choice. Let $R$ be an arbitrary relation. There are two cases to consider: \subparagraph{Case 1}% Suppose $\ran{R} = \emptyset$. Then $R = \emptyset$. Thus the function $\emptyset \subseteq R$ satisfies $\dom{\emptyset} = \dom{R}$. \subparagraph{Case 2}% Suppose $\ran{R} \neq \emptyset$. Let $I = \dom{R}$ and define $H \colon I \rightarrow \{\ran{R}\}$ as $H(i) = \ran{R}$ for all $i \in I$. By the \nameref{ref:axiom-of-choice-2}, $\bigtimes_{i \in I} H(i) \neq \emptyset$. By definition of the \nameref{ref:cartesian-product}, there exists some function $f$ such that $\dom{f} = I$ and $(\forall i \in I) f(i) \in H(i) = \ran{R}$. Thus $\dom{f} = \dom{R}$ and $f \subseteq R$ as desired. \paragraph{Conclusion}% The above cases are exhaustive and yield the same conclusion: for any relation $R$ there exists a function $f \subseteq R$ such that $\dom{f} = \dom{R}$. \end{proof} \subsection{\verified{Exercise 3.32a}}% \hyperlabel{sub:exercise-3.32-a} Show that $R$ is symmetric iff $R^{-1} \subseteq R$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_32\_a} \begin{proof} \paragraph{($\Rightarrow$)}% Suppose $R$ is \nameref{ref:symmetric}. Let $\tuple{x, y} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, $\tuple{y, x} \in R$. By symmetry, $\tuple{x, y} \in R$. Thus $R^{-1} \subseteq R$. \paragraph{($\Leftarrow$)}% Suppose $R^{-1} \subseteq R$. Let $\tuple{x, y} \in R$. By definition of the \nameref{ref:inverse} of a set, $\tuple{y, x} \in R^{-1}$. Since $R^{-1} \subseteq R$, $\tuple{y, x} \in R$. Therefore $\tuple{x, y}$ and $\tuple{y, x}$ are both in $R$. In other words, $R$ is symmetric. \end{proof} \subsection{\verified{Exercise 3.32b}}% \hyperlabel{sub:exercise-3.32-b} Show that $R$ is transitive iff $R \circ R \subseteq R$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_32\_b} \begin{proof} \paragraph{($\Rightarrow$)}% Suppose $R$ is \nameref{ref:transitive}. Let $\tuple{x, y} \in R \circ R$. By definition of the \nameref{ref:composition} of a set, there exists some $t$ such that $xRt \land tRy$. That is, $\tuple{x, t} \in R$ and $\tuple{t, y} \in R$. Since $R$ is transitive, it follows $\tuple{x, y} \in R$. \paragraph{($\Leftarrow$)}% Suppose $R \circ R \subseteq R$. Let $\tuple{x, y} \tuple{y, z} \in R$. By definition of the \nameref{ref:composition} of a set, $$R \circ R = \{\tuple{u, v} \mid \exists t(uRt \land tRv)\}.$$ Then $\tuple{x, z} \in R \circ R$. Since $R \circ R \subseteq R$, it follows $\tuple{x, z} \in R$. Thus $R$ is transitive. \end{proof} \subsection{\verified{Exercise 3.33}}% \hyperlabel{sub:exercise-3.33} Show that $R$ is a symmetric and transitive relation iff $R = R^{-1} \circ R$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_33} \begin{proof} By definition of the \nameref{ref:inverse} and \nameref{ref:composition} of sets, \begin{align} R^{-1} \circ R & = \{ (u, v) \mid \exists t(uRt \land tR^{-1}v) \} \nonumber \\ & = \{ (u, v) \mid \exists t(uRt \land vRt) \}. \hyperlabel{sub:exercise-3.33-eq1} \end{align} \paragraph{($\Rightarrow$)}% Suppose $R$ is symmetric and transitive. We now show that $R \subseteq R^{-1} \circ R$ and $R^{-1} \circ R \subseteq R$. \subparagraph{($\subseteq$)}% Let $\tuple{x, y} \in R$. Since $R$ is symmetric, $\tuple{y, x} \in R$. Since $R$ is transitive, $\tuple{x, x} \in R$. Then there exists a $t$ such that $\tuple{x, t} \in R$ and $\tuple{y, t} \in R$, namely $t = x$. By \eqref{sub:exercise-3.33-eq1}, $\tuple{x, y} \in R^{-1} \circ R$. \subparagraph{($\supseteq$)}% Let $\tuple{x, y} \in R^{-1} \circ R$. By \eqref{sub:exercise-3.33-eq1}, there exists some $t$ such that $\tuple{x, t} \in R$ and $\tuple{y, t} \in R$. But $R$ is symmetric meaning $\tuple{t, y} \in R$. Since $R$ is transitive, it follows $\tuple{x, y} \in R$. \paragraph{($\Leftarrow$)}% Suppose $R = R^{-1} \circ R$. We prove that (i) $R$ is symmetric and (ii) $R$ is transitive. \subparagraph{(i)}% \hyperlabel{spar:exercise-3.33-i} First we note that $R$ is equal to its inverse: \begin{align} R^{-1} & = (R^{-1} \circ R)^{-1} \nonumber \\ & = R^{-1} \circ (R^{-1})^{-1} & \textref{sub:theorem-3i} \nonumber \\ & = R^{-1} \circ R & \textref{sub:theorem-3e} \nonumber \\ & = R \hyperlabel{sub:exercise-3.33-eq2}. \end{align} Now let $\tuple{x, y} \in R$. By \eqref{sub:exercise-3.33-eq2} $\tuple{x, y} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, $\tuple{y, x} \in R$. Thus $R$ is symmetric. \subparagraph{(ii)}% Let $\tuple{x, y}, \tuple{y, z} \in R$. By \nameref{spar:exercise-3.33-i}, $R$ is symmetric. Thus $\tuple{z, y} \in R$. By \eqref{sub:exercise-3.33-eq1}, it follows $\tuple{x, z} \in R^{-1} \circ R$. Since $R^{-1} \circ R = R$, it follows $\tuple{x, z} \in R$. Thus $R$ is transitive. \end{proof} \subsection{\verified{Exercise 3.34}}% \hyperlabel{sub:exercise-3.34} Assume that $\mathscr{A}$ is a nonempty set, every member of which is a transitive relation. \begin{enumerate}[(a)] \item Is the set $\bigcap{\mathscr{A}}$ a transitive relation? \item Is $\bigcup{\mathscr{A}}$ a transitive relation? \end{enumerate} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_34\_a} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_34\_b} \begin{proof} \paragraph{(a)}% Because $\mathscr{A} \neq \emptyset$, $\bigcap{\mathscr{A}}$ is well-defined. We prove that $\bigcap{\mathscr{A}}$ is a transitive relation. Let $\tuple{x, y}, \tuple{y, z} \in \bigcap{\mathscr{A}}$. Then forall $A$ in $\mathscr{A}$, it follows $\tuple{x, y}, \tuple{y, z} \in A$. Since $A$ is transitive, it follows $\tuple{x, z} \in A$. Since this holds for all $A \in \mathscr{A}$, it follows that $\tuple{x, z} \in A$ as well. Thus $\bigcap{\mathscr{A}}$ is transitive. \paragraph{(b)}% We show that $\bigcup{\mathscr{A}}$ is not necessarily transitive with a counterexample. Suppose $$\mathscr{A} = \{ \{\tuple{1, 2}, \tuple{2, 3}, \tuple{1, 3}\}, \{\tuple{2, 1}\} \}.$$ Notice that the two members of $\mathscr{A}$ are transitive relations. Now $$\bigcup{\mathscr{A}} = \{ \tuple{1, 2}, \tuple{2, 3}, \tuple{1, 3}, \tuple{2, 1}, \}.$$ But the above cannot be transitive, for $\tuple{1, 2}$ and $\tuple{2, 1}$ are members of the set, but $\tuple{1, 1}$ is not. \end{proof} \subsection{\verified{Exercise 3.35}}% \hyperlabel{sub:exercise-3.35} Show that for any $R$ and $x$, we have $[x]_R = \img{R}{\{x\}}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_35} \begin{proof} Let $R$ and $x$ be arbitrary sets. Then \begin{align*} [x]_R & = \{t \mid xRt\} \\ & = \{t \mid (\exists u \in \{x\})uRt\} \\ & = \img{R}{\{x\}}. \end{align*} \end{proof} \subsection{\verified{Exercise 3.36}}% \hyperlabel{sub:exercise-3.36} Assume that $f \colon A \rightarrow B$ and that $R$ is an equivalence relation on $B$. Define $Q$ to be the set \begin{equation} \hyperlabel{sub:exercise-3.36-eq1} \{\tuple{x, y} \in A \times A \mid \tuple{f(x), f(y)} \in R\}. \end{equation} Show that $Q$ is an equivalence relation on $A$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_36} \begin{proof} We prove that (i) $Q$ is \nameref{ref:reflexive} on $A$, (ii) $Q$ is \nameref{ref:symmetric}, and (iii) $Q$ is \nameref{ref:transitive}. \paragraph{(i)}% Let $x \in A$. By hypothesis, $f(x) \in B$. Since $R$ is an equivalence relation on $B$, $R$ is reflexive on $B$. Thus $\tuple{f(x), f(x)} \in R$. But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{x, x} \in Q$. Thus $Q$ is reflexive on $A$. \paragraph{(ii)}% Let $\tuple{x, y} \in Q$. By \eqref{sub:exercise-3.36-eq1}, $\tuple{f(x), f(y)} \in R$. Since $R$ is an equivalence relation on $B$, $R$ is symmetric. Thus $\tuple{f(y), f(x)} \in R$. But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{y, x} \in Q$. Thus $Q$ is symmetric. \paragraph{(iii)}% Let $\tuple{x, y}, \tuple{y, z} \in Q$. By \eqref{sub:exercise-3.36-eq1}, $\tuple{f(x), f(y)}, \tuple{f(y), f(z)} \in R$. Since $R$ is an equivalence relation on $B$, $R$ is transitive. Thus $\tuple{f(x), f(z)} \in R$. But then \eqref{sub:exercise-3.36-eq1} implies $\tuple{x, z} \in Q$. Thus $Q$ is transitive. \end{proof} \subsection{\verified{Exercise 3.37}}% \hyperlabel{sub:exercise-3.37} Assume that $\Pi$ is a partition of a set $A$. Define the relation $R_\Pi$ as follows: \begin{equation} \hyperlabel{sub:exercise-3.37-eq1} xR_{\Pi}y \iff (\exists B \in \Pi)(x \in B \land y \in B). \end{equation} Show that $R_\Pi$ is an equivalence relation on $A$. (This is a formalized version of the discussion at the beginning of this section.) \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_37} \begin{proof} We prove that (i) $R_\Pi$ is \nameref{ref:reflexive} on $B$, (ii) $R_\Pi$ is \nameref{ref:symmetric}, and (iii) $R_\Pi$ is \nameref{ref:transitive}. \paragraph{(i)}% Let $x \in A$. By definition of a \nameref{ref:partition}, there exists some nonempty set $B \in \Pi$ such that $x \in B$. Thus $(\exists B \in \Pi)(x \in B \land x \in B)$. By \eqref{sub:exercise-3.37-eq1}, $\tuple{x, x} \in R_\Pi$. Therefore $R_\Pi$ is reflexive on $A$. \paragraph{(ii)}% Let $\tuple{x, y} \in R_\Pi$. By \eqref{sub:exercise-3.37-eq1}, there exists some $B \in \Pi$ such that $x \in B \land y \in B$. But then $y \in B \land x \in B$. Thus $\tuple{y, x} \in R_\Pi$. In other words, $R_\Pi$ is symmetric. \paragraph{(iii)}% Let $\tuple{x, y}, \tuple{y, z} \in R_\Pi$. By \eqref{sub:exercise-3.37-eq1}, there exists some $B_1 \in \Pi$ such that $x \in B_1 \land y \in B_1$. Likewise there exists some $B_2 \in \Pi$ such that $y \in B_2 \land z \in B_2$. But $\Pi$ is a \nameref{ref:partition} meaning $y \in B_1$ and $y \in B_2$ if $B_1 = B_2$. Therefore $x \in B_1 \land z \in B_1$ and $\tuple{x, z} \in R_\Pi$. In other words, $R_\Pi$ is transitive. \end{proof} \subsection{\verified{Exercise 3.38}}% \hyperlabel{sub:exercise-3.38} \nameref{sub:theorem-3p} shows that $A / R$ is a partition of $A$ whenever $R$ is an equivalence relation on $A$. Show that if we start with the equivalence relation $R_\Pi$ of the preceding exercise, then the partition $A / R_\Pi$ is just $\Pi$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_38} \begin{proof} By definition, \begin{equation} \hyperlabel{sub:exercise-3.38-eq1} R_\Pi = \{ (x, y) \mid (\exists B \in \Pi)(x \in B \land y \in B) \}. \end{equation} We prove that $A / R_\Pi = \Pi$. By the \nameref{ref:extensionality-axiom}, these two sets are equal when $$B \in A / R_\Pi \iff B \in \Pi.$$ We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Suppose $B \in A / R_\Pi$. By \nameref{sub:exercise-3.37}, $R_\Pi$ is an equivalence class. Then, by definition of a \nameref{ref:quotient-set}, $$A / R_\Pi = \{[x]_{R_\Pi} \mid x \in A\},$$ whose members are the \nameref{ref:equivalence-class}es. Thus there exists some $x \in A$ such that $B = [x]_{R_\Pi}$. By definition of a \nameref{ref:partition}, there exists a unique set $B' \in \Pi$ containing $x$. Thus it suffices to prove that $B = B'$, for then $B = [x]_{R_\Pi} = B'$ is a member of $\Pi$ as desired. We proceed by extensionality again; that is, we show $$y \in B \iff y \in B'.$$ \subparagraph{($\rightarrow$)}% Suppose $y \in B$. Then \begin{align*} y & \in B = [x]_{R_\Pi} \\ & = \{t \mid \tuple{x, t} \in R_\Pi\} \\ & = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\}. & \eqref{sub:exercise-3.38-eq1} \end{align*} Thus there exists some $B_1 \in \Pi$ such that $x \in B_1$ and $y \in B_1$. By definition of a \nameref{ref:partition}, $B_1 \in \Pi$ is the unique member of $\Pi$ containing $y$. Thus $B_1 = B'$ meaning $y \in B'$ as desired. \subparagraph{($\leftarrow$)}% Suppose $y \in B'$. By construction, $x \in B'$. Then there exists a set $B_1$ such that $x \in B_1$ and $y \in B_1$, namely $B'$. Therefore \begin{align*} y & \in \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\ & = \{t \mid \tuple{x, t} \in R_\Pi\} \\ & = [x]_{R_\Pi} = B. \end{align*} \subparagraph{Conclusion}% By the \nameref{ref:extensionality-axiom}, it follows $B = B'$. Since $B' \in P$, it also follows $B \in P$. \paragraph{($\Leftarrow$)}% Let $B \in \Pi$. By definition of a \nameref{ref:partition}, $B$ is nonempty. Let $x \in B$. By definition of a set, $B = \{t \mid x \in B \land t \in B\}$. By definition of a \nameref{ref:partition}, every member of $B$ must belong to only $B$ (i.e. no other sets in the partition). Thus we can equivalently write \begin{align*} B & = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\ & = \{ t \mid \tuple{x, t} \in R_\Pi \} \\ & = [x]_{R_\Pi}. \end{align*} Therefore $B \in A / R_{\Pi}$. \end{proof} \subsection{\verified{Exercise 3.39}}% \hyperlabel{sub:exercise-3.39} Assume that we start with an equivalence relation $R$ on $A$ and define $\Pi$ to be the partition $A / R$. Show that $R_\Pi$, as defined in Exercise 37, is just $R$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_39} \begin{proof} By definition, \begin{equation} \hyperlabel{sub:exercise-3.39-eq1} R_\Pi = \{ (x, y) \mid (\exists B \in \Pi)(x \in B \land y \in B) \}. \end{equation} We prove that $R_\Pi = R$. By the \nameref{ref:extensionality-axiom}, these two sets are equal when $$(x, y) \in R_\Pi \iff (x, y) \in R.$$ We prove both directions of this biconditional. \paragraph{($\Rightarrow$)}% Let $(x, y) \in R_\Pi$. By \eqref{sub:exercise-3.39-eq1}, there exists some $B \in \Pi$ such that $x \in B$ and $y \in B$. Since $\Pi = A / R = \{[x]_R \mid x \in A\}$, there must exist some $z \in A$ such that $B = [z]_R$. By definition of an \nameref{ref:equivalence-class}, $x \in [z]_R$ implies that $zRx$ and $y \in [z]_R$ implies $zRy$. Since $R$ is \nameref{ref:symmetric}, $xRz$. Since $R$ is \nameref{ref:transitive}, $xRy$. \paragraph{($\Leftarrow$)}% Let $(x, y) \in R$. By definition of an \nameref{ref:equivalence-class}, $x \in [x]_R$ and $y \in [x]_R$. Note also that $[x]_R \in A / R = \Pi$. Thus there exists some $B \in \Pi$ such that $x \in B$ and $y \in B$, namely $B = [x]_R$. By \eqref{sub:exercise-3.39-eq1}, $(x, y) \in R_\Pi$. \end{proof} \subsection{\unverified{Exercise 3.40}}% \hyperlabel{sub:exercise-3.40} Define an equivalence relation $R$ on the set $P$ of positive integers by $$mRn \iff m \text{ and } n \text{ have the same number of prime factors}.$$ Is there a function $f \colon P / R \rightarrow P / R$ such that $f([n]_R) = [3n]_R$ for each $n$? \begin{proof} Define $g \colon P \rightarrow P$ as $g(x) = 3x$ for all $x \in P$. We first show that $g$ is \nameref{ref:compatible} with $R$. Let $m, n \in P$ such that $mRn$. Then $m$ and $n$ have the same prime factors. Then $3m$ has one additional prime factor than $m$, namely $3$. Likewise $3n$ has one additional prime factor than $n$, also $3$. Thus $(3m)R(3n)$, i.e. $g$ is compatible with $R$. By \nameref{sub:theorem-3q}, it follows there exists a unique function $f \colon P / R \rightarrow P / R$ such that $f([n]_R) = [g(n)]_R = [3n]_R$ as expected. \end{proof} \subsection{\unverified{Exercise 3.41}}% \hyperlabel{sub:exercise-3.41} Let $\mathbb{R}$ be the set of real numbers and define the relation $Q$ on $\mathbb{R} \times \mathbb{R}$ by $\tuple{u, v}Q\tuple{x, y}$ iff $u + y = x + v$. \subsubsection{\verified{Exercise 3.41a}}% \hyperlabel{ssub:exercise-3.41-a} Show that $Q$ is an equivalence relation on $\mathbb{R} \times \mathbb{R}$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_41\_a} \begin{proof} We show (i) $Q$ is \nameref{ref:reflexive} on $\mathbb{R} \times \mathbb{R}$, (ii) $Q$ is \nameref{ref:symmetric}, and (iii) $Q$ is \nameref{ref:transitive}. \paragraph{(i)}% Let $\tuple{x, y} \in R \times R$. Since $x + y = x + y$, it immediately follows $\tuple{x, y}Q\tuple{x, y}$. Thus $Q$ is reflexive on $\mathbb{R}$. \paragraph{(ii)}% Let $\tuple{\tuple{u, v}, \tuple{x, y}} \in Q$. Then $u + y = x + v$. Likewise, $x + v = u + y$. This immediately implies that $\tuple{\tuple{x, y}, \tuple{u, v}} \in Q$. Thus $Q$ is symmetric. \paragraph{(iii)}% Let $\tuple{\tuple{u, v}, \tuple{x, y}} \in Q$ and $\tuple{\tuple{x, y}, \tuple{a, b}} \in Q$. Then $u + y = x + v$ and $x + b = a + y$. Rearranging terms, we have $u - v = x - y$ and $x - y = a - b$. Thus $u - v = a - b$. Rearranging terms once more yields $u + b = a + v$. Thus $\tuple{\tuple{u, v}, \tuple{a, b}} \in Q$. Therefore $Q$ is transitive. \end{proof} \subsubsection{\unverified{Exercise 3.41b}}% \hyperlabel{ssub:exercise-3.41-b} Is there a function $G \colon (\mathbb{R} \times \mathbb{R}) / Q \rightarrow (\mathbb{R} \times \mathbb{R}) / Q$ satisfying the equation \begin{equation} \hyperlabel{ssub:exercise-3.41-b-eq1} G([\tuple{x, y}]_Q) = [\tuple{x + 2y, y + 2x}]_Q? \end{equation} \begin{proof} Let $f \colon \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ be given by $f(\tuple{x, y}) = \tuple{x + 2y, y + 2x}$. We show (i) that $f$ is \nameref{ref:compatible} with $Q$ and then (ii) there exists such a function satisfying \eqref{ssub:exercise-3.41-b-eq1}. \paragraph{(i)}% \hyperlabel{par:exercise-3.41-b-i} Let $\tuple{u, v}, \tuple{x, y} \in \mathbb{R} \times \mathbb{R}$ such that $\tuple{u, v} Q \tuple{x, y}$. Thus \begin{equation} \hyperlabel{ssub:exercise-3.41-b-eq2} u + y = x + v \end{equation} Next consider \begin{align*} f(\tuple{u, v}) & = \tuple{u + 2v, v + 2u}, \\ f(\tuple{x, y}) & = \tuple{x + 2y, y + 2x}. \end{align*} Then \begin{align*} u + y & = x + v \\ \iff 3u + 3y & = 3x + 3v \\ \iff (u + y) + (2u + 2y) & = (x + v) + (2x + 2v) \\ \iff (x + v) + (2u + 2y) & = (u + y) + (2x + 2v) & \eqref{ssub:exercise-3.41-b-eq2} \\ \iff (x + 2y) + (v + 2u) & = (u + 2v) + (y + 2x) \\ \iff (u + 2v) + (y + 2x) & = (x + 2y) + (v + 2u). \end{align*} This last equality shows $f(\tuple{u, v}) \,Q\, f(\tuple{x, y})$. Thus $f$ is compatible with $Q$. \paragraph{(ii)}% By \nameref{par:exercise-3.41-b-i} and \nameref{sub:theorem-3q}, there exists a unique $G \colon (\mathbb{R} \times \mathbb{R}) / Q \rightarrow (\mathbb{R} \times \mathbb{R}) / Q$ satisfying \eqref{ssub:exercise-3.41-b-eq1}. \end{proof} \subsection{\unverified{Exercise 3.42}}% \hyperlabel{sub:exercise-3.42} State precisely the "analogous results" mentioned in \nameref{sub:theorem-3q}. (This will require extending the concept of compatibility in a suitable way.) \linedivider \begin{theorem}[3.42] A function $F \colon A \times A \rightarrow A$ is said to be \textbf{compatible} with relation $R$ if and only if for all $x_1, y_1, x_2, y_2 \in A$, $$x_1Rx_2 \land y_1Ry_2 \Rightarrow F(x_1, y_1)RF(x_2, y_2).$$ \noindent Assume that $R$ is an equivalence relation on $A$ and that $F \colon A \times A \rightarrow A$. If $F$ is compatible with $R$, then there exists a unique function $\hat{F} \colon A / R \times A / R \rightarrow A / R$ such that \begin{equation} \hyperlabel{sub:exercise-3.42-eq1} \hat{F}([x]_R, [y]_R) = [F(x, y)]_R \end{equation} for all $x, y \in A$. If $F$ is not compatible with $R$, then no such $\hat{F}$ exists. \end{theorem} \begin{proof} Let $R$ be an equivalence relation on $A$ and $F \colon A \times A \rightarrow A$ be compatible with $R$. Define relation $\hat{F}$ to be $$\hat{F} = \{\tuple{[x]_R, [y]_R, [F(x, y)]_R} \mid x, y \in A\}.$$ By construction, $\dom{\hat{F}} = A / R \times A / R$ and $\ran{\hat{F}} \subseteq A / R$. All that remains is proving (i) $\hat{F}$ is single-valued and (ii) $\hat{F}$ is unique. \paragraph{(i)}% Let $[x_1]_R, [y_1]_R, [x_2]_R, [y_2]_R \in \dom{\hat{F}}$ such that \begin{equation} \hyperlabel{par:theorem-3q-i-eq1} \tuple{[x_1]_R, [y_1]_R} = \tuple{[x_2]_R, [y_2]_R}. \end{equation} By definition of $\hat{F}$, \begin{align*} \tuple{[x_1]_R, [y_1]_R, [F(x_1, y_1)]_R} & \in \hat{F} \\ \tuple{[x_2]_R, [y_2]_R, [F(x_2, y_2)]_R} & \in \hat{F}. \end{align*} By \eqref{par:theorem-3q-i-eq1}, $[x_1]_R = [x_2]_R$ and $[y_1]_R = [y_2]_R$. Then \nameref{sub:lemma-3n} implies $x_1Rx_2$ and $y_1Ry_2$ respectively. Since $F$ is compatible, it follows $F(x_1, y_1)RF(x_2, y_2)$. Another application of \nameref{sub:lemma-3n} implies that $[F(x_1, y_1)]_R = [F(x_2, y_2)]_R$. Thus $\hat{F}$ is single-valued, i.e. a function. \paragraph{(ii)}% Suppose there exists another function, say $\hat{G}$, that satisfies \eqref{sub:exercise-3.42-eq1}. That is, $$\hat{G}([x]_R, [y]_R) = [F(x, y)]_R \quad\text{for all } x, y \in A.$$ Let $x, y \in A$. Then $\hat{G}([x]_R, [y]_R) = [F(x, y)]_R$ and $\hat{F}([x]_R, [y]_R) = [F(x, y)]_R$. Since this holds for all $x, y \in A$, $\hat{F}$ and $\hat{G}$ agree on all members of $A / R \times A / R$. Hence, by the \nameref{ref:extensionality-axiom}, $\hat{F} = \hat{G}$. \suitdivider \noindent Suppose $F$ is not compatible with $R$. Then there exists some $x_1, y_1, x_2, y_2 \in A$ such that $x_1Rx_2$ and $y_1Ry_2$ but $\neg F(x_1, y_1)RF(x_2, y_2)$. By \nameref{sub:lemma-3n}, $[x_1]_R = [x_2]_R$ and $[y_1]_R = [y_2]_R$. For the sake of contradiction, suppose a function $\hat{F}$ exists satisfying \eqref{sub:exercise-3.42-eq1}. Then $\hat{F}([x_1]_R, [y_1]_R) = \hat{F}([x_2]_R, [y_2]_R)$ meaning $[F(x_1, y_1)]_R = [F(x_2, y_2)]_R$. Then \nameref{sub:lemma-3n} implies $F(x_1, y_1)RF(x_2, y_2)$, a contradiction. Therefore our original hypothesis must be incorrect. That is, there is no incompatible function $\hat{F}$ satisfying \eqref{sub:exercise-3.42-eq1}. \end{proof} \subsection{\verified{Exercise 3.43}}% \hyperlabel{sub:exercise-3.43} Assume that $R$ is a linear ordering on a set $A$. Show that $R^{-1}$ is also a linear ordering on $A$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_43} \begin{proof} Assume that $R$ is a \nameref{ref:linear-ordering} on a set $A$. Then $R$ is \nameref{ref:transitive} and \nameref{ref:trichotomous}. We show that (i) $R^{-1}$ is transitive and (ii) $R^{-1}$ is trichotomous. \paragraph{(i)}% \hyperlabel{par:exercise-3.43-i} Let $\tuple{x, y}, \tuple{y, z} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, $\tuple{y, x}$, $\tuple{z, y} \in R$. Since $R$ is transitive, it must be that $\tuple{z, x} \in R$. Then $\tuple{x, z} \in R^{-1}$. Thus $R^{-1}$ is transitive. \paragraph{(ii)}% \hyperlabel{par:exercise-3.43-ii} Let $x, y \in A$. Since $R$ is trichotomous on $A$, it follows that exactly one of the following conditions hold: $$xRy, \quad x = y, \quad yRx.$$ By definition of the \nameref{ref:inverse} of a set, the above possibilities are equivalently expressed as $$yR^{-1}x, \quad x = y, \quad xR^{-1}y.$$ Thus $R^{-1}$ is trichotomous. \paragraph{Conclusion}% Since $R^{-1}$ is transitive by \nameref{par:exercise-3.43-i} and trichotomous by \nameref{par:exercise-3.43-ii}, it follows $R^{-1}$ is a linear ordering on $A$. \end{proof} \subsection{\verified{Exercise 3.44}}% \hyperlabel{sub:exercise-3.44} Assume that $<$ is a linear ordering on a set $A$. Assume that $f \colon A \rightarrow A$ and that $f$ has the property that whenever $x < y$, then $f(x) < f(y)$. Show that $f$ is one-to-one and that whenever $f(x) < f(y)$, then $x < y$. \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_44\_i} \code{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_44\_ii} \begin{proof} We show that (i) $f$ is one-to-one and (ii) whenever $f(x) < f(y)$, then $x < y$. \paragraph{(i)}% Let $y \in \ran{f}$. By definition of the \nameref{ref:range} of a set, there exists some $x_1 \in A$ such that $f(x_1) = y$. Suppose there exists some $x_2 \in A$ such that $f(x_2) = y$. We prove $f$ is one-to-one by showing $x_1 = x_2$. Because $<$ is a linear ordering on $A$, there exist three cases to consider: \subparagraph{Case 1}% Assume $x_1 < x_2$. By hypothesis, $f$ is monotonic. Thus $f(x_1) < f(x_2)$. But $<$ is a trichotomous relation meaning it is not possible for \textit{both} $f(x) = f(y)$ and $f(x) < f(y)$. Thus our original assumption must be wrong. \subparagraph{Case 2}% Assume $x_1 = x_2$. Then we are immediately finished. \subparagraph{Case 3}% Assume $x_1 > x_2$. By hypothesis, $f$ is monotonic. Thus $f(x_1) > f(x_2)$. But $<$ is a trichotomous relation meaning it is not possible for \textit{both} $f(x) = f(y)$ and $f(x) > f(y)$. Thus our original assumption must be wrong. \subparagraph{Conclusion}% Since the above cases are exhaustive, the only possibility is $x_1 = x_2$. Thus $f$ is one-to-one. \paragraph{(ii)}% Suppose $f(x) < f(y)$. There are three cases to consider: \subparagraph{Case 1}% Assume $x < y$. Then we are immediately finished. \subparagraph{Case 2}% Assume $x = y$. Then $f(x) = f(y)$. But $<$ is a trichotomous relation meaning it is not possible for \textit{both} $f(x) = f(y)$ and $f(x) < f(y)$. Thus our original assumption must be wrong. \subparagraph{Case 3}% Assume $x > y$. By hypothesis, $f$ is monotonic. Thus $f(x) > f(y)$. But $<$ is a trichotomous relation meaning it is not possible for \textit{both} $f(x) < f(y)$ and $f(x) > f(y)$. Thus our original assumption must be wrong. \subparagraph{Conclusion}% Since the above cases are exhaustive, the only possibility is $x < y$. \end{proof} \subsection{\verified{Exercise 3.45}}% \hyperlabel{sub:exercise-3.45} Assume that $<_A$ and $<_B$ are linear orderings on $A$ and $B$, respectively. Define the binary relation $<_L$ on the Cartesian product $A \times B$ by: $$\tuple{a_1, b_1} <_L \tuple{a_2, b_2} \quad\text{iff}\quad \text{either } a_1 <_A a_2 \text{ or } (a_1 = a_2 \land b_1 <_B b_2).$$ Show that $<_L$ is a linear ordering on $A \times B$. (The relation $<_L$ is called \textit{lexicographic} ordering, being the ordering used in making dictionaries.) \code*{Bookshelf/Enderton/Set/Chapter\_3} {Enderton.Set.Chapter\_3.exercise\_3\_45} \begin{proof} We show that $<_L$ is (i) \nameref{ref:transitive} and (ii) \nameref{ref:trichotomous} on $A \times B$. \paragraph{(i)}% Let $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$ and $\tuple{a_2, b_2} <_L \tuple{a_3, b_3}$. Then either $a_1 <_A a_2$ or $a_1 = a_2 \land b_1 <_B b_2$. Likewise, either $a_2 <_A a_3$ or $a_2 = a_3 \land b_2 <_B b_3$. We consider each combination of cases in turn: \subparagraph{Case 1}% Suppose $a_1 <_A a_2$ and $a_2 <_A a_3$. Since $<_A$ is a linear ordering, it follows $<_A$ is transitive. Thus $a_1 <_A a_3$. Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 2}% Suppose $a_1 <_A a_2$, $a_2 = a_3$, and $b_2 <_B b_3$. Then $a_1 < a_3$. Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 3}% Suppose $a_1 = a_2$, $b_1 <_B b_2$, and $a_2 <_A a_3$. Then $a_1 <_A a_3$. Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Case 4}% Suppose $a_1 = a_2$, $b_1 <_B b_2$, $a_2 = a_3$, and $b_2 <_B b_3$. Then $a_1 = a_3$. Since $<_B$ is a linear ordering, it follows $<_B$ is transitive. Thus $b_1 <_B b_3$. Therefore $\tuple{a_1, b_2} <_L \tuple{a_3, b_3}$. \subparagraph{Conclusion}% These four cases are exhaustive and each conclude that $<_L$ is transitive. \paragraph{(ii)}% Let $\tuple{a_1, b_1}, \tuple{a_2, b_2} \in A \times B$. Because $<_A$ and $<_B$ are linear orderings on $A$ and $B$ respectively, it follows $<_A$ and $<_B$ are both trichotomous on their respective sets. Thus exactly one of $$a_1 <_A a_2, \quad a_1 = a_2, \quad a_2 <_A a_1$$ and $$b_1 <_B b_2, \quad b_1 = b_2, \quad b_2 <_B b_1$$ holds. There are three cases we examine: \subparagraph{Case 1}% \hyperlabel{spar:exercise-3.45-ii-case-1} Suppose $a_1 <_A a_2$. Then $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$. This is trivially the only possible relationship between the ordered pairs. \subparagraph{Case 2}% Suppose $a_1 = a_2$. If $b_1 <_B b_2$, then $\tuple{a_1, b_1} <_L \tuple{a_2, b_2}$ is the only possibility. If $b_1 = b_2$, then $\tuple{a_1, b_1} = \tuple{a_2, b_2}$ is the only possibility. If $b_2 <_B b_1$, then $\tuple{a_2, b_2} <_L \tuple{a_1, b_1}$ is the only possibility. \subparagraph{Case 3}% Suppose $a_2 <_A a_1$. This case is analagous to \eqref{spar:exercise-3.45-ii-case-1}. \subparagraph{Conclusion}% In each of the above cases, we are always left with exactly one of $$\tuple{a_1, b_1} <_L \tuple{a_2, b_2}, \quad \tuple{a_1, b_1} = \tuple{a_2, b_2}, \quad \tuple{a_2, b_2} <_L \tuple{a_1, b_1}.$$ Thus $<_L$ is trichotomous. \end{proof} \chapter{Natural Numbers}% \hyperlabel{chap:natural-numbers} \section{Inductive Sets}% \hyperlabel{sec:inductive-sets} \subsection{\unverified{Theorem 4A}}% \hyperlabel{sub:theorem-4a} \begin{theorem}[4A] There is a set whose members are exactly the natural numbers. \end{theorem} \begin{proof} By the \nameref{ref:infinity-axiom}, there exists an \nameref{ref:inductive-set} $A$. By the \nameref{ref:subset-axioms}, there exists a set $B$ such that $$x \in B \iff x \in A \land \left[\forall C, (\emptyset \in C \land (\forall c \in C) c^+ \in C) \Rightarrow x \in C\right].$$ In other words, $x \in B$ if and only if $x \in A$ and $x$ is a natural number. Thus $B$ is the set whose members are exactly the natural numbers. \end{proof} \subsection{\unverified{Theorem 4B}}% \hyperlabel{sub:theorem-4b} \begin{theorem}[4B] $\omega$ is inductive, and is a subset of every other inductive set. \end{theorem} \begin{proof} $\omega$ denotes the set of \nameref{ref:natural-number}s. We show $\omega$ is an \nameref{ref:inductive-set} by proving (i) $\emptyset \in \omega$ and (ii) $\omega$ is closed under \nameref{ref:successor}. \paragraph{(i)}% \hyperlabel{par:theorem-4b-i} By definition, $\emptyset$ is a member of every inductive set. Thus $\emptyset$ is a natural number, i.e. a member of $\omega$. \paragraph{(ii)}% \hyperlabel{par:theorem-4b-ii} Let $n \in \omega$. That is, let $n$ be a natural number. By definition, $n$ is a member of every inductive set. By definition of an inductive set, $n^+$ is then a member of every inductive set as well. Thus $n^+$ is a natural number, i.e. $n^+ \in \omega$. \paragraph{Conclusion}% By \nameref{par:theorem-4b-i} and \nameref{par:theorem-4b-ii}, it follows $\omega$ is inductive. It follows immediately from the definition of a natural number that $\omega$ is a subset of every other inductive set. \end{proof} \subsection{\verified{Theorem 4C}}% \hyperlabel{sub:theorem-4c} \begin{theorem}[4C] Every natural number except $0$ is the successor of some natural number. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4c} \begin{proof} Let $T = \{n \mid n = 0 \lor (\exists m) n = m^+\}$. It trivially follows that $\emptyset \in T$. Let $x \in T$. Then $x^+ \in T$ since $(\exists m) x^+ = m^+$, namely $m = x$. Therefore $T$ is inductive. By \nameref{sub:theorem-4b}, $\omega$ is a subset of $T$. Thus every natural number satisfies the condition written in $T$'s definition. In other words, every natural number except $0$ is the successor of some natural number. \end{proof} \section{Peano's Postulates}% \hyperlabel{sec:peanos-postulates} \subsection{\unverified{Theorem 4E}}% \hyperlabel{sub:theorem-4e} \begin{theorem}[4E] For a transitive set $a$, $$\bigcup \left(a^+\right) = a.$$ \end{theorem} \begin{proof} Let $a$ be a \nameref{ref:transitive-set}. We show that \begin{equation} \hyperlabel{sub:theorem-4e-eq1} x \in \bigcup \left(a^+\right) \iff x \in a. \end{equation} \paragraph{($\Rightarrow$)}% Suppose $x \in \bigcup \left(a^+\right)$. By definition of \nameref{ref:successor}, $x \in \bigcup \left(a \cup \{a\}\right)$. Then there exists some $b \in a \cup \{a\}$ such that $x \in b$. There are two cases to consider: \subparagraph{Case 1}% Suppose $b \in a$. By definition of a transitive set, $x \in b \in a$ means $x \in a$. \subparagraph{Case 2}% Suppose $b \in \{a\}$. Then $b = a$ and immediately $x \in b = a$. \paragraph{($\Leftarrow$)}% Suppose $x \in a$. Then immediately $x \in a \cup \{a\}$. Thus there exists some $b$ such that $b \in a \cup \{a\}$ and $x \in b$, namely $b = \{a\}$. Thus $x \in \bigcup \left(a^+\right)$. \paragraph{Conclusion}% We have shown both sides of \eqref{sub:theorem-4e-eq1} holds. By the \nameref{ref:extensionality-axiom}, $\bigcup \left(a^+\right) = a$. \end{proof} \subsection{\unverified{Theorem 4F}}% \hyperlabel{sub:theorem-4f} \begin{theorem}[4F] Every natural number is a transitive set. \end{theorem} \begin{proof} Let $T = \{n \in \omega \mid n \text{ is a transitive set}\}$. We (i) prove that $T$ is an \nameref{ref:inductive-set} and then (ii) every natural number is a transitive set. \paragraph{(i)}% \hyperlabel{par:theorem-4f-i} First, $\emptyset \in T$ since it vacuously holds that a member of a member of $\emptyset$ is itself a member of $\emptyset$. Next, let $n \in T$ and consider whether $n^+ \in T$. Since $n$ is a transitive set, \nameref{sub:theorem-4e} implies $\bigcup \left(n^+\right) = n$. But $n \subseteq n^+ = n \cup \{n\}$. Thus $\bigcup \left(n^+\right) \subseteq n+$, i.e. $n^+$ is a transitive set. Therefore $n^+ \in T$. Hence $T$ is inductive. \paragraph{(ii)}% Notice $T \subseteq \omega$. By \nameref{par:theorem-4f-i} and \nameref{sub:theorem-4b}, $T = \omega$. Thus every natural number is a transitive set. \end{proof} \subsection{\verified{Theorem 4D}}% \hyperlabel{sub:theorem-4d} \begin{theorem}[4D] $\langle \omega, \sigma, 0 \rangle$ is a Peano system. \end{theorem} \code{Common/Set/Peano}{Peano.instSystemNatUnivSuccOfNatInstOfNatNat} \begin{note} This theorem depends on \nameref{sub:theorem-4e} and \nameref{sub:theorem-4f}. \end{note} \begin{proof} Note $\sigma$ is defined as $\sigma = \{\tuple{n, n^+} \mid n \in \omega\}$. To prove $\langle \omega, \sigma, 0 \rangle$ is a \nameref{ref:peano-system}, we must show that (i) $0 \not\in \ran{S}$, (ii) $\sigma$ is one-to-one, and (iii) every subset $A$ of $\omega$ containing $0$ and closed under $\sigma$ is $\omega$ itself. \paragraph{(i)}% This follows immediately from \nameref{sub:theorem-4c}. \paragraph{(ii)}% Let $m, n \in \omega$ and suppose $m^+ = n^+$. Then $\bigcup \left(m^+\right) = \bigcup \left(n^+\right)$. By \nameref{sub:theorem-4f}, every natural number is a \nameref{ref:transitive-set}. Therefore, by \nameref{sub:theorem-4e}, $$\bigcup \left(m^+\right) = m = \bigcup \left(n^+\right) = n.$$ \paragraph{(iii)}% This follows immediately from \nameref{sub:theorem-4b}. \end{proof} \subsection{\unverified{Theorem 4G}}% \hyperlabel{sub:theorem-4g} \begin{theorem}[4G] The set $\omega$ is a transitive set. \end{theorem} \begin{proof} Let $T = \{n \in \omega \mid \forall t \in n, t \in \omega\}$. We prove that (i) $T$ is an \nameref{ref:inductive-set} and then (ii) every member of a natural number is itself a natural number. \paragraph{(i)}% \hyperlabel{par:theorem-4g-i} First, it vacuously holds that $\emptyset \in T$. Next, let $n \in T$. We must prove that $n^+ \in T$ as well. By definition of the \nameref{ref:successor}, $n^+ = n \cup \{n\}$. That is, either $n^+ = n$ or $n^+ = \{n\}$. If the former, then every member of $n^+$ must be a natural number since this already holds for $n$. If the latter, the only member of $n^+$ is $n$ which is, by definition of $T$, a natural number. Thus $n^+ \in T$. We conclude that $T$ is an inductive set. \paragraph{(ii)}% Since $T \subseteq \omega$, \nameref{par:theorem-4g-i} and \nameref{sub:theorem-4b} implies $T = \omega$. Thus the member of every natural number is itself a natural number. In other words, $\bigcup \omega \subseteq \omega$. Therefore $\omega$ is indeed a \nameref{ref:transitive-set}. \end{proof} \section{Recursion on \texorpdfstring{$\omega$}{the Natural Numbers}}% \hyperlabel{sec:recursion-natural-numbers} \subsection{\unverified{% Recursion Theorem on \texorpdfstring{$\omega$}{the Natural Numbers}}}% \hyperlabel{sub:recursion-theorem-natural-numbers} \begin{theorem} Let $A$ be a set, $a \in A$, and $F \colon A \rightarrow A$. Then there exists a unique function $h \colon \omega \rightarrow A$ such that $$h(0) = a,$$ and for every $n \in \omega$, $$h(n^+) = F(h(n)).$$ \end{theorem} \begin{note} This proof was written a few days after reading Enderton's proof as a means of ensuring I remember the main arguments. \end{note} \begin{proof} Define set \begin{align*} H = \{ v \mid & v \text{ is a function with } \\ & \text{(a) } \dom{v} \subseteq \omega, \\ & \text{(b) } \ran{v} \subseteq A, \\ & \text{(c) if } 0 \in \dom{v}, \text{ then } v(0) = a, \text{ and} \\ & \text{(d) if } n^+ \in \dom{v}, \text{ then } n \in \dom{v} \text{ and } v(n^+) = F(v(n)) \}. \end{align*} Define a function satisfying properties (a)-(d) above as \textit{acceptable}. That is, $H$ is the set of all acceptable functions. Define $h = \bigcup H$. We prove that (i) $h$ is a \nameref{ref:function}, (ii) $h \in H$, (iii) $\dom{h} = \omega$, and (iv) $h$ is unique. \paragraph{(i)}% \hyperlabel{par:recursion-theorem-natural-numbers-i} We prove that $h$ is a function. Consider set $$S = \{x \in \omega \mid h(x) = y \text{ for at most one value } y\}.$$ We show (1) that $0 \in S$ and (2) if $n \in S$ then $n^+ \in S$. \subparagraph{(1)}% \hyperlabel{spar:recursion-theorem-natural-numbers-i-1} Suppose $0 \in \dom{h}$. By construction, there must exist some $y_1 \in A$ and acceptable function $v_1$ such that $v_1(0) = y_1$. Suppose there also exists a $y_2 \in A$ and acceptable function $v_2$ such that $v_2(0) = y_2$. By property (c), $v_1(0) = a$ and $v_2(0) = a$. Thus $y_1 = a = y_2$ and $h(0) = a$. Therefore $0 \in S$. \subparagraph{(2)}% \hyperlabel{spar:recursion-theorem-natural-numbers-i-2} Suppose $n$ and $n^+$ are members of $\dom{h}$. By construction, there must exist some $y_1 \in A$ and acceptable function $v_1$ such that $v_1(n^+) = y_1$. Suppose there also exists a $y_2 \in A$ and acceptable function $v_2$ such that $v_2(n^+) = y_2$. By property (d), it follows $n \in \dom{v_1}$, $n \in \dom{v_2}$, $v_1(n^+) = F(v_1(n))$, and $v_2(n^+) = F(v_2(n))$. But $n \in S$ meaning there is at most one value $y$ such that $v_1(n) = y = v_2(n)$. Thus $F(v_1(n)) = F(y) = F(v_2(n))$ and $h(n^+) = F(y)$. Therefore $n^+ \in S$. \subparagraph{Subconclusion}% By \nameref{spar:recursion-theorem-natural-numbers-i-1} and \nameref{spar:recursion-theorem-natural-numbers-i-2}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Since $S = \omega$, it follows $h$ has at most one value for every $x \in \omega$. In other words, $h$ is a function. \paragraph{(ii)}% \hyperlabel{par:recursion-theorem-natural-numbers-ii} We now prove $h \in H$, i.e. $h$ is an acceptable function. It trivially holds that $\dom{h} \subseteq \omega$ and $\ran{h} \subseteq A$. Thus we are left with proving properties (c) and (d). \subparagraph{(c)}% Note $\{\tuple{0, a}\}$ is an acceptable function. Thus $\tuple{0, a} \in h$. By \nameref{par:recursion-theorem-natural-numbers-i}, $h$ is a function. Therefore $a$ is the only value $h(0)$ takes on. \subparagraph{(d)}% Suppose $n^+ \in \dom{h}$. Then there exists some acceptable function $v$ such that $v(n^+) = h(n^+)$. By definition of acceptable, $\tuple{n, v(n)} \in v$. Since \nameref{par:recursion-theorem-natural-numbers-i} indicates $h$ is a function, $n \in \dom{h}$ and $h(n) = v(n)$. Also by definition of acceptable, $v(n^+) = F(v(n))$. Therefore $$h(n^+) = v(n^+) = F(v(n)) = F(h(n)).$$ Hence $h \in H$. \paragraph{(iii)}% \hyperlabel{par:recursion-theorem-natural-numbers-iii} We now prove that $\dom{h} = \omega$. We show that (1) $0 \in \dom{h}$ and (2) if $n \in \dom{h}$ then $n^+ \in \dom{h}$. \subparagraph{(1)}% \hyperlabel{spar:recursion-theorem-natural-numbers-iii-1} We note that $\{\tuple{0, a}\}$ is an acceptable function. By construction of $h$, $0 \in \dom{h}$. \subparagraph{(2)}% \hyperlabel{spar:recursion-theorem-natural-numbers-iii-2} Suppose $n \in \dom{h}$. Since $n \in \dom{h}$ there exists an acceptable function $v$ with $n \in \dom{v}$. Define $$v' = v \cup \{\tuple{n^+, F(v(n))}\}.$$ We prove that $v'$ is acceptable: \begin{enumerate}[(a)] \item It trivially holds that $\dom{v'} \subseteq \omega$. \item It trivially holds that $\ran{v'} \subseteq A$. \item If $0 \in \dom{v'}$, then $v'(0) = v(0) = a$. \item Suppose $m^+ \in \dom{v'}$ for some $m \in \omega$. If $m^+ = n^+$, then $m = n$ and $v'(m^+) = F(v(m))$ by construction. If $m^+ \neq n^+$, then $m^+ \in \dom{v}$. Since $v$ is an acceptable function, $v'(m^+) = v(m^+) = F(v(m)) = F(v(m^+))$. \end{enumerate} Since $v'$ is acceptable, $n^+ \in \dom{h}$. \subparagraph{Subconclusion}% By \nameref{spar:recursion-theorem-natural-numbers-iii-1} and \nameref{spar:recursion-theorem-natural-numbers-iii-2}, $\dom{h}$ is an inductive set. \nameref{sub:theorem-4b} implies $\dom{h} = \omega$. \paragraph{(iv)}% \hyperlabel{par:recursion-theorem-natural-numbers-iv} We now prove $h$ is a unique function. Let $h_1$ and $h_2$ both satisfy the conclusion of the theorem. Define $$S = \{n \in \omega \mid h_1(n) = h_2(n)\}.$$ It suffices to prove $S$ is an inductive set, for \nameref{sub:theorem-4b} would then imply $S = \omega$, i.e. $h_1$ and $h_2$ agree on all of $\omega$. By definition of an acceptable function, $h_1(0) = a = h_2(0)$ meaning $0 \in S$. Next, suppose $n \in S$. By \nameref{par:recursion-theorem-natural-numbers-iii}, it follows $n^+$ in $\dom{h_1}$ and $\dom{h_2}$. Since both $h_1$ and $h_2$ are acceptable, $h_1(n^+) = F(h_1(n))$ and $h_2(n^+) = F(h_2(n))$. Since $n \in S$, $h_1(n) = h_2(n)$. Therefore $h_1$ and $h_2$ coincide with input $n^+$. Thus $n^+ \in S$. Hence $S$ is an inductive set. \paragraph{Conclusion}% By \nameref{par:recursion-theorem-natural-numbers-i}, \nameref{par:recursion-theorem-natural-numbers-iii}, and \nameref{par:recursion-theorem-natural-numbers-iv}, it follows $h$ is a unique function mapping $\omega$ into $A$. \nameref{par:recursion-theorem-natural-numbers-ii} shows $h$ satisfies the desired conditions. \end{proof} \subsection{\verified{Theorem 4H}}% \hyperlabel{sub:theorem-4h} \begin{theorem}[4H] Let $\langle N, S, e \rangle$ be a Peano system. Then $\langle \omega, \sigma, 0 \rangle$ is isomorphic to $\langle N, S, e \rangle$, i.e., there is a function $h$ mapping $\omega$ one-to-one onto $N$ in a way that preserves the successor operation $$h(\sigma(n)) = S(h(n))$$ and the zero element $$h(0) = e.$$ \end{theorem} \code{Common/Set/Peano} {Peano.nat\_isomorphism} \begin{proof} Let $\langle N, S, e \rangle$ be a \nameref{ref:peano-system}. By the \nameref{sub:recursion-theorem-natural-numbers}, there exists a unique function $h \colon \omega \rightarrow N$ such that $h(0) = e$ and for every $n \in \omega$, $h(n^+) = h(\sigma(n)) = S(h(n))$. All that remains is proving $h$ is one-to-one and onto. \suitdivider \noindent We first show $h$ is one-to-one by induction. Define $$S = \{n \in \omega \mid \forall m, h(m) = h(n) \Rightarrow m = n\}.$$ We show that (i) $0 \in S$ and (ii) if $n \in S$, then $n^+ \in S$. Afterward we show (iii) that $h$ is one-to-one. \paragraph{(i)}% \hyperlabel{par:theorem-4h-i} Let $m \in \omega$ such that $h(m) = h(0) = e$. Suppose $m \neq 0$. Then \nameref{sub:theorem-4c} indicates there exists some natural number $p$ such that $p^+ = m$. But then $h(m) = h(p^+) = S(h(p)) = e$. By definition of a Peano system, $e \not\in \ran{S}$. This is a contradiction. Hence $m = 0$, i.e. $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4h-ii} Suppose $n \in S$. Let $m \in \omega$ such that $h(m) = h(n^+) = S(h(n))$. There are two cases to consider: \subparagraph{Case 1}% Suppose $m = 0$. Then $h(0) = e = S(h(n))$. But, by definition of a Peano system, $e \not\in \ran{S}$. Thus we have a contradiction. \subparagraph{Case 2}% Suppose $m \neq 0$. Then \nameref{sub:theorem-4c} indicates there exists some natural number $p$ such that $p^+ = m$. Then $$S(h(n)) = h(m) = h(p^+) = S(h(p)).$$ By definition of a Peano system, $S$ is one-to-one. Therefore $h(n) = h(p)$. Since $n \in S$, it follows $n = p$. Therefore $n^+ = p^+ = m$. \subparagraph{Subconclusion}% The above two cases are exhaustive. Hence $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:theorem-4h-i} and \nameref{par:theorem-4h-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all natural numbers $m, n \in \omega$, if $h(m) = h(n)$, then $m = n$. In other words, $f$ is one-to-one. \suitdivider \noindent We next show that $\ran{h} = N$. By the Peano posulate, every subset $A$ of $N$ containing $e$ and closed under $S$ is $N$ itself. Thus it suffices to prove that (i) $e \in \ran{h}$ and (ii) $\ran{h}$ is closed under $S$. \paragraph{(i)}% That is, $h(0) = e$ by definition. Thus $e \in \ran{h}$. \paragraph{(ii)}% Let $y \in \ran{h}$. Then there exists some $n \in \omega$ such that $h(n) = y$. By definition of $h$, $h(n^+) = S(h(n)) = S(y)$. Therefore $S(y) \in \ran{h}$. Since this holds for any $y \in \ran{h}$, $\ran{h}$ is closed under $S$. \end{proof} \section{Arithmetic}% \hyperlabel{sec:arithmetic} \subsection{\verified{Theorem 4I}} \hyperlabel{sub:theorem-4i} \begin{theorem}[4I] For \nameref{ref:natural-number}s $m$ and $n$, \begin{align} m + 0 & = m, \hyperlabel{sub:theorem-4i-eq1} \\ m + n^+ & = (m + n)^+. \hyperlabel{sub:theorem-4i-eq2} \end{align} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4i} \lean{Init/Data/Nat/Basic}{Nat.add\_zero} \lean{Init/Prelude}{Nat.add} \begin{proof} \paragraph{\eqref{sub:theorem-4i-eq1}}% Let $m$ be a \nameref{ref:natural-number}. By definition of \nameref{ref:addition}, $m + 0 = A_m(0)$. By definition of $A_m$, $A_m(0) = m$. Thus $m + 0 = m$. \paragraph{\eqref{sub:theorem-4i-eq2}}% Let $m$ and $n$ be natural numbers. By definition of \nameref{ref:addition}, $$m + n^+ = A_m(n^+) = A_m(n)^+ = (m + n)^+.$$ \end{proof} \subsection{\verified{Theorem 4J}} \hyperlabel{sub:theorem-4j} \begin{theorem}[4J] For natural numbers $m$ and $n$, \begin{align} m \cdot 0 & = 0, \hyperlabel{sub:theorem-4j-eq1} \\ m \cdot n^+ & = m \cdot n + m. \hyperlabel{sub:theorem-4j-eq2} \end{align} \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4j} \lean{Init/Data/Nat/Basic}{Nat.mul\_zero} \lean{Init/Prelude}{Nat.mul} \begin{proof} \paragraph{\eqref{sub:theorem-4j-eq1}}% Let $m$ be a \nameref{ref:natural-number}. By definition of \nameref{ref:multiplication}, $$m \cdot 0 = M_m(0) = 0.$$ \paragraph{\eqref{sub:theorem-4j-eq2}}% Let $m$ and $n$ be natural numbers. By definition of \nameref{ref:multiplication}, $$m \cdot n^+ = M_m(n^+) = M_m(n) + m = m \cdot n + m.$$ \end{proof} \subsection{\verified{Left Additive Identity}}% \hyperlabel{sub:left-additive-identity} \begin{lemma} For all $n \in \omega$, $A_0(n) = n$. In other words, $$0 + n = n.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.left\_additive\_identity} \lean{Init/Data/Nat/Basic}{Nat.zero\_add} \begin{proof} Let $S = \{n \in \omega \mid 0 + n = n\}$. We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:left-additive-identity-i} By \nameref{sub:theorem-4i}, $0 + 0 = 0$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:left-additive-identity-ii} Suppose $n \in S$. By \nameref{sub:theorem-4i}, $0 + n^+ = (0 + n)^+$. Since $n \in S$, $0 + n = n$ which in turn implies that $(0 + n)^+ = n^+$. Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:left-additive-identity-i} and \nameref{par:left-additive-identity-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, $0 + n = n$. \end{proof} \subsection{\verified{Successor Commutativity}}% \hyperlabel{sub:successor-commutativity} \begin{lemma} For all $m, n \in \omega$, $A_{m^+}(n) = A_m(n^+)$. In other words, $$m^+ + n = m + n^+.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.lemma\_2} \lean{Std/Data/Nat/Lemmas}{Nat.succ\_add\_eq\_succ\_add} \begin{proof} Let $m \in \omega$ and define $$S = \{n \in \omega \mid m^+ + n = m + n^+\}.$$ We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:lemma-2-i} By \nameref{sub:theorem-4i}, $m^+ + 0 = m^+$. Likewise, $m + 0^+ = (m + 0)^+ = m^+$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:lemma-2-ii} Suppose $n \in S$. By \nameref{sub:theorem-4i}, $m^+ + n^+ = (m^+ + n)^+$. Since $n \in S$, $m^+ + n = m + n^+$. Therefore $(m^+ + n)^+ = (m + n^+)^+ = m + n^{++}$. Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:lemma-2-i} and \nameref{par:lemma-2-ii}, $S$ is inductive. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, $m^+ + n = m + n^+$. \end{proof} \subsection{\verified{Theorem 4K-1}}% \hyperlabel{sub:theorem-4k-1} \begin{theorem}[4K-1] Associative law for \nameref{ref:addition}. For $m, n, p \in \omega$, $$m + (n + p) = (m + n) + p.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4k\_1} \lean{Init/Data/Nat/Basic}{Nat.add\_assoc} \begin{proof} Fix $n, p \in \omega$ and define \begin{equation} \hyperlabel{sub:theorem-4k-1-eq1} S = \{m \in \omega \mid m + (n + p) = (m + n) + p\}. \end{equation} We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. Afterward we show that (iii) the associative law for addition holds. \paragraph{(i)}% \hyperlabel{par:theorem-4k-1-i} By \nameref{sub:left-additive-identity}, $$0 + (n + p) = n + p = (0 + n) + p.$$ Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4k-1-ii} Suppose $m \in S$. Then \begin{align*} m^+ + (n + p) & = m + (n + p)^+ & \textref{sub:successor-commutativity} \\ & = (m + (n + p))^+ & \textref{sub:theorem-4i} \\ & = ((m + n) + p)^+ & \eqref{sub:theorem-4k-1-eq1} \\ & = (m + n) + p^+ & \textref{sub:theorem-4i} \\ & = (m + n)^+ + p & \textref{sub:successor-commutativity} \\ & = (m + n^+) + p & \textref{sub:theorem-4i} \\ & = (m^+ + n) + p. & \textref{sub:successor-commutativity} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% \hyperlabel{par:theorem-4k-1-iii} By \nameref{par:theorem-4k-1-i} and \nameref{par:theorem-4k-1-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n, p \in \omega$, $m + (n + p) = (m + n) + p$. \end{proof} \subsection{\verified{Theorem 4K-2}}% \hyperlabel{sub:theorem-4k-2} \begin{theorem}[4K-2] Commutative law for \nameref{ref:addition}. For $m, n \in \omega$, $$m + n = n + m.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4k\_2} \lean{Init/Data/Nat/Basic}{Nat.add\_comm} \begin{proof} Fix $n \in \omega$ and define \begin{equation} \hyperlabel{sub:theorem-4k-2-eq1} S = \{m \in \omega \mid m + n = n + m\}. \end{equation} We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. Afterward we show that (iii) the commutative law for addition holds. \paragraph{(i)}% \hyperlabel{par:theorem-4k-2-i} By definition of \nameref{ref:addition} and \nameref{sub:left-additive-identity}, $$0 + n = n = n + 0.$$ Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4k-2-ii} Suppose $m \in S$. Then \begin{align*} m^+ + n & = m + n^+ & \textref{sub:successor-commutativity} \\ & = (m + n)^+ & \textref{sub:theorem-4i} \\ & = (n + m)^+ & \eqref{sub:theorem-4k-2-eq1} \\ & = n + m^+. & \textref{sub:theorem-4i} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:theorem-4k-2-i} and \nameref{par:theorem-4k-2-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n \in \omega$, $m + n = n + m$. \end{proof} \subsection{\verified{Zero Multiplicand}}% \hyperlabel{sub:zero-multiplicand} \begin{lemma} For all $n \in \omega$, $M_0(n) = 0$. In other words, $$0 \cdot n = 0.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.zero\_multiplicand} \lean{Init/Data/Nat/Basic}{Nat.zero\_mul} \begin{proof} Define \begin{equation} \hyperlabel{sub:zero-multiplicand-eq1} S = \{n \in \omega \mid 0 \cdot n = 0\}. \end{equation} We prove that (i) $0 \in S$ and (ii) that if $n \in S$, then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:zero-multiplicand-i} By \nameref{sub:theorem-4j}, $0 \cdot 0 = 0$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:zero-multiplicand-ii} Suppose $n \in S$. Then \begin{align*} 0 \cdot n^+ & = 0 \cdot n + 0 & \textref{sub:theorem-4j} \\ & = 0 \cdot n & \textref{sub:theorem-4i} \\ & = 0. & \eqref{sub:zero-multiplicand-eq1} \end{align*} Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:zero-multiplicand-i} and \nameref{par:zero-multiplicand-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, $0 \cdot n = 0$. \end{proof} \subsection{\verified{Successor Distribution}}% \hyperlabel{sub:successor-distribution} \begin{lemma} For all $m, n \in \omega$, $M_{m^+}(n) = M_m(n) + n$. In other words, $$m^+ \cdot n = m \cdot n + n.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.succ\_distrib} \lean{Init/Data/Nat/Basic}{Nat.succ\_mul} \begin{proof} Let $m \in \omega$ and define \begin{equation} \hyperlabel{sub:successor-distribution-eq1} S = \{n \in \omega \mid m^+ \cdot n = m \cdot n + n\}. \end{equation} We prove that (i) $0 \in S$ and (ii) that if $n \in S$, then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:successor-distribution-i} By \nameref{sub:theorem-4j}, $m^+ \cdot 0 = 0$. Likewise, by \nameref{sub:theorem-4i}, $m \cdot 0 + 0 = 0$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:successor-distribution-ii} Suppose $n \in S$. Then \begin{align*} m^+ \cdot n^+ & = m^+ \cdot n + m^+ & \textref{sub:theorem-4j} \\ & = (m \cdot n + n) + m^+ & \eqref{sub:successor-distribution-eq1} \\ & = m \cdot n + (n + m^+) & \textref{sub:theorem-4k-1} \\ & = m \cdot n + (n^+ + m) & \textref{sub:successor-commutativity} \\ & = m \cdot n + (m + n^+) & \textref{sub:theorem-4k-2} \\ & = (m \cdot n + m) + n^+ & \textref{sub:theorem-4k-1} \\ & = m \cdot n^+ + n^+. & \textref{sub:theorem-4j} \end{align*} Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:successor-distribution-i} and \nameref{par:successor-distribution-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n \in \omega$, $m^+ \cdot n = m \cdot n + n$. \end{proof} \subsection{\verified{Theorem 4K-3}} \hyperlabel{sub:theorem-4k-3} \begin{theorem}[4K-3] Distributive law. For $m, n, p \in \omega$, $$m \cdot (n + p) = m \cdot n + m \cdot p.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4k\_3} \lean{Init/Data/Nat/Basic}{Nat.left\_distrib} \begin{proof} Fix $n, p \in \omega$ and define \begin{equation} \hyperlabel{sub:theorem-4k-3-eq1} S = \{m \in \omega \mid m \cdot (n + p) = m \cdot n + m \cdot p\}. \end{equation} We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. Afterward we show that (iii) the distributive law holds. \paragraph{(i)}% \hyperlabel{par:theorem-4k-3-i} By definition of \nameref{ref:multiplication} and \nameref{ref:addition}, \begin{align*} 0 \cdot (n + p) & = 0 & \textref{sub:zero-multiplicand} \\ & = 0 + 0 & \textref{ref:addition} \\ & = 0 \cdot n + 0 \cdot p. & \textref{sub:zero-multiplicand} \end{align*} Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4k-3-ii} Suppose $m \in S$. By definition of \nameref{ref:multiplication} and \nameref{ref:addition}, \begin{align*} m^+ \cdot (n + p) & = m \cdot (n + p) + (n + p) & \textref{sub:successor-distribution} \\ & = m \cdot (n + p) + n + p & \textref{sub:theorem-4k-1} \\ & = m \cdot n + m \cdot p + n + p & \eqref{sub:theorem-4k-3-eq1} \\ & = m \cdot n + n + m \cdot p + p & \textref{sub:theorem-4k-2} \\ & = m^+ \cdot n + m^+ \cdot p. & \textref{sub:successor-distribution} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:theorem-4k-3-i} and \nameref{par:theorem-4k-3-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n, p \in \omega$, $m \cdot (n + p) = m \cdot n + m \cdot p$. \end{proof} \subsection{\verified{Successor Identity}}% \hyperlabel{sub:successor-identity} \begin{lemma} For all $m \in \omega$, $A_m(1) = m^+$. In other words, $$m + 1 = m^+.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.succ\_identity} \lean{Std/Data/Nat/Lemmas}{Nat.succ\_eq\_one\_add} \begin{proof} Let \begin{equation} \hyperlabel{sub:successor-identity-eq1} S = \{m \in \omega \mid m + 1 = m^+\}. \end{equation} We prove that (i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:successor-identity-i} By \nameref{sub:left-additive-identity}, $0 + 1 = 1$. By definition of the \nameref{ref:successor}, $0^+ = \emptyset \cup \{\emptyset\} = 1$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:successor-identity-ii} Let $m \in S$. Then \begin{align*} m^+ + 1 & = m + 1^+ & \textref{sub:successor-commutativity} \\ & = (m + 1)^+ & \textref{sub:theorem-4i} \\ & = (m^+)^+. & \eqref{sub:successor-identity-eq1} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:successor-identity-i} and \nameref{par:successor-identity-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $m \in \omega$, $m + 1 = m^+$. \end{proof} \subsection{\verified{Right Multiplicative Identity}}% \hyperlabel{sub:right-multiplicative-identity} \begin{lemma} For all $m \in \omega$, $M_m(1) = m$. In other words, $$m \cdot 1 = m.$$ \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.right\_mul\_id} \lean{Init/Data/Nat/Basic}{Nat.mul\_one} \begin{proof} Let \begin{equation} \hyperlabel{sub:right-multiplicative-identity-eq1} S = \{m \in \omega \mid m \cdot 1 = m\}. \end{equation} We prove that (i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:right-multiplicative-identity-i} By \nameref{sub:zero-multiplicand}, $0 \cdot 1 = 0$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:right-multiplicative-identity-ii} Suppose $m \in S$. Then \begin{align*} m^+ \cdot 1 & = m \cdot 1 + 1 & \textref{sub:successor-distribution} \\ & = m + 1 & \eqref{sub:right-multiplicative-identity-eq1} \\ & = m^+. & \textref{sub:successor-identity} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:right-multiplicative-identity-i} and \nameref{par:right-multiplicative-identity-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m \in \omega$, $m \cdot 1 = m$. \end{proof} \subsection{\verified{Theorem 4K-5}} \hyperlabel{sub:theorem-4k-5} \begin{theorem}[4K-5] Commutative law for \nameref{ref:multiplication}. For $m, n \in \omega$, $$m \cdot n = n \cdot m.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4k\_5} \lean{Init/Data/Nat/Basic}{Nat.mul\_comm} \begin{note} We prove commutativity before associativity, though Enderton orders these two properties in the opposite direction. \end{note} \begin{proof} Fix $n \in \omega$ and define \begin{equation} \hyperlabel{sub:theorem-4k-5-eq1} S = \{m \in \omega \mid m \cdot n = n \cdot m\}. \end{equation} We show that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. Afterward we show that (iii) the commutative law for multiplication holds. \paragraph{(i)}% \hyperlabel{par:theorem-4k-5-i} By \nameref{sub:theorem-4j} and \nameref{sub:zero-multiplicand}, $$0 \cdot n = 0 = n \cdot 0.$$ Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4k-5-ii} Suppose $m \in S$. Then \begin{align*} m^+ \cdot n & = m \cdot n + n & \textref{sub:successor-distribution} \\ & = n \cdot m + n & \eqref{sub:theorem-4k-5-eq1} \\ & = n \cdot m + n \cdot 1 & \textref{sub:right-multiplicative-identity} \\ & = n \cdot (m + 1) & \textref{sub:theorem-4k-3} \\ & = n \cdot m^+. & \textref{sub:successor-identity} \end{align*} Thus $m^+ \in S$. \paragraph{(iii)}% \hyperlabel{par:theorem-4k-5-iii} By \nameref{par:theorem-4k-5-i} and \nameref{par:theorem-4k-5-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n \in \omega$, $m \cdot n = n \cdot m$. \end{proof} \subsection{\verified{Theorem 4K-4}}% \hyperlabel{sub:theorem-4k-4} \begin{theorem}[4K-4] Associative law for \nameref{ref:multiplication}. For $m, n, p \in \omega$, $$m \cdot (n \cdot p) = (m \cdot n) \cdot p.$$ \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4k\_4} \lean{Init/Data/Nat/Basic}{Nat.mul\_assoc} \begin{proof} Fix $m, n \in \omega$ and define \begin{equation} \hyperlabel{sub:theorem-4k-4-eq1} S = \{p \in \omega \mid m \cdot (n \cdot p) = (m \cdot n) \cdot p\}. \end{equation} We show that (i) $0 \in S$ and (ii) if $p \in S$ then $p^+ \in S$. Afterward we show that (iii) the associative law for multiplication holds. \paragraph{(i)}% \hyperlabel{par:theorem-4k-4-i} By \nameref{sub:theorem-4j}, $$m \cdot (n \cdot 0) = 0 = (m \cdot n) \cdot 0.$$ Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:theorem-4k-4-ii} Suppose $p \in S$. Then \begin{align*} m \cdot (n \cdot p^+) & = m \cdot (n \cdot p + n) & \textref{sub:theorem-4j} \\ & = m \cdot (n \cdot p) + m \cdot n & \textref{sub:theorem-4k-3} \\ & = (m \cdot n) \cdot p + m \cdot n & \eqref{sub:theorem-4k-4-eq1} \\ & = p \cdot (m \cdot n) + m \cdot n & \textref{sub:theorem-4k-5} \\ & = p^+ \cdot (m \cdot n) & \textref{sub:successor-distribution} \\ & = (m \cdot n) \cdot p^+ & \textref{sub:theorem-4k-5} \end{align*} Thus $p^+ \in S$. \paragraph{(iii)}% By \nameref{par:theorem-4k-4-i} and \nameref{par:theorem-4k-4-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n, p \in \omega$, $m \cdot (n \cdot p) = (m \cdot n) \cdot p$. \end{proof} \section{Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}% \hyperlabel{sec:ordering-natural-numbers} \subsection{\unverified{Ordering on Successor}}% \hyperlabel{sub:ordering-successor} \begin{lemma} Let $m, n \in \omega$. Then $m < n^+ \iff m \leq n$. \end{lemma} \lean{Std/Data/Nat/Lemmas}{Nat.lt\_succ} \begin{proof} Let $m, n \in \omega$. By \nameref{ref:ordering-natural-numbers}, \begin{align*} m < n^+ & \iff m \in n^+ \\ & \iff m \in n \cup \{n\} & \textref{ref:successor} \\ & \iff m \in n \lor m \in \{n\} \\ & \iff m \in n \lor m = n \\ & \iff m \leq n. \end{align*} \end{proof} \subsection{\unverified{Members of Natural Numbers}}% \hyperlabel{sub:members-natural-numbers} \begin{lemma} Every \nameref{ref:natural-number} is the set of all smaller natural numbers. \end{lemma} \begin{proof} Let $n \in \omega$. Consider $m \in n$. By \nameref{sub:theorem-4b}, $\omega$ is a \nameref{ref:transitive-set}. Thus $m \in n$ implies $m \in \omega$. Thus $m \in n \iff m \in \omega \land m \in n$. \end{proof} \subsection{\unverified{Lemma 4L(a)}}% \hyperlabel{sub:lemma-4l-a} \begin{lemma}[4L(a)] For any \nameref{ref:natural-number}s $m$ and $n$, $$m \in n \iff m^+ \in n^+.$$ \end{lemma} \lean{Std/Data/Nat/Lemmas}{Nat.succ\_lt\_succ\_iff} \begin{note} Here I referred to Enderton's proof in the forward direction. \end{note} \begin{proof} Let $m$ and $n$ be \nameref{ref:natural-number}s. \paragraph{($\Rightarrow$)}% Define $$S = \{n \in \omega \mid (\forall m \in n) m^+ \in n^+\}.$$ We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterwards we show that (iii) the forward direction of the stated biconditional holds. \subparagraph{(i)}% \hyperlabel{spar:lemma-4l-a-i} $0 \in S$ vacuously. That is, there are no members of $0 = \emptyset$ by definition. \subparagraph{(ii)}% \hyperlabel{spar:lemma-4l-a-ii} Suppose $n \in S$. We need to show for all $m \in n^+$, $m^+ \in n^{++}$. Let $m \in n^+ = n \cup \{n\}$. Then $m \in n$ or $m \in \{n\}$. If $m \in n$, then $n \in S$ implies $m^+ \in n^+ \in n^{++}$. By \nameref{sub:theorem-4f}, every natural number is a \nameref{ref:transitive-set}. Therefore $m^+ \in n^{++}$. On the other hand, if $m \in \{n\}$, then $m = n$. Since $n^+ \in n^{++}$, it immediately follows $m^+ \in n^{++}$. Hence $n^+ \in S$. \subparagraph{(iii)}% By \nameref{spar:lemma-4l-a-i} and \nameref{spar:lemma-4l-a-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, $m \in n \Rightarrow m^+ \in n^+$. \paragraph{($\Leftarrow$)}% Suppose $m^+ \in n^+$. The definition of \nameref{ref:successor} immediately implies that $m \in m^+$. Likewise, $m^+ \in n^+$ implies $m^+ \in n$ or $m^+ = n$. If the latter, $m \in n$ immediately follows. If the former, we note $n$ is a transitive set by \nameref{sub:theorem-4f}. Thus $m \in m^+ \in n$ implies $m \in n$. \end{proof} \subsection{\verified{Lemma 4L(b)}}% \hyperlabel{sub:lemma-4l-b} \begin{lemma}[4L(b)] No \nameref{ref:natural-number} is a member of itself. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4l\_b} \lean{Init/Prelude}{Nat.lt\_irrefl} \begin{proof} Define \begin{equation} \hyperlabel{sub:lemma-4l-b-eq1} S = \{n \in \omega \mid n \not\in n\}. \end{equation} We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:lemma-4l-b-i} By definition, $0 = \emptyset$. It obviously holds that $\emptyset \not\in \emptyset$ since $\emptyset$, by definition, has no members. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:lemma-4l-b-ii} Suppose $n \in S$. By \eqref{sub:lemma-4l-b-eq1}, $n \not\in n$. By \nameref{sub:lemma-4l-a}, it follows $n^+ \not\in n^+$. Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:lemma-4l-b-i} and \nameref{par:lemma-4l-b-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, $n \not\in n$. \end{proof} \subsection{\verified{\texorpdfstring{$0$}{Zero} is the Least Natural Number}}% \hyperlabel{sub:zero-least-natural-number} \begin{lemma} For every \nameref{ref:natural-number} $n \neq 0$, $0 \in n$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.zero\_least\_nat} \lean{Std/Data/Nat/Init/Lemmas}{Nat.pos\_of\_ne\_zero} \begin{proof} Let $$S = \{n \in \omega \mid n = 0 \lor 0 \in n\}.$$ We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:zero-least-natural-number-i} This trivially holds by definition of $S$. \paragraph{(ii)}% \hyperlabel{par:zero-least-natural-number-ii} Suppose $n \in S$. By definition of the \nameref{ref:successor} function, $n^+ = n \cup \{n\}$. Thus $n \in n^+$. By \nameref{sub:theorem-4f}, $n^+$ is a \nameref{ref:transitive-set}. Since $0 \in n$ and $n \in n^+$, it follows $0 \in n^+$. Thus $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:zero-least-natural-number-i} and \nameref{par:zero-least-natural-number-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $n \in \omega$, either $n = 0$ or $0 \in n$. \end{proof} \subsection{\unverified{% Trichotomy Law for \texorpdfstring{$\omega$}{Natural Numbers}}}% \hyperlabel{sub:trichotomy-law-natural-numbers} \begin{theorem} For any \nameref{ref:natural-number}s $m$ and $n$, exactly one of the three conditions $$m \in n, \quad m = n, \quad n \in m$$ holds. \end{theorem} \lean{Mathlib/Order/RelClasses}{IsAsymm} \lean{Mathlib/Init/Algebra/Classes}{IsTrichotomous} \begin{proof} Let $n \in \omega$ and define \begin{equation} \hyperlabel{sub:trichotomy-law-natural-numbers-eq1} S = \{m \in \omega \mid m \in n \lor m = n \lor n \in m\}. \end{equation} We prove that (i) $0 \in S$ and (ii) if $m \in S$ then $m^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:trichotomy-law-natural-numbers-i} If $n = 0$, then it trivially follows $0 \in S$. Otherwise \nameref{sub:zero-least-natural-number} implies $0 \in n$. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:trichotomy-law-natural-numbers-ii} Suppose $m \in S$. By \eqref{sub:trichotomy-law-natural-numbers-eq1}, there are three cases to consider: \subparagraph{Case 1}% Suppose $m \in n$. By \nameref{sub:lemma-4l-a}, $m^+ \in n^+ = n \cup \{n\}$. By definition of the \nameref{ref:successor}, $m^+ \in n$ or $m^+ = n$. Either way, $m^+ \in S$. \subparagraph{Case 2}% Suppose $m = n$. Since $m \in m^+$, it follows $n \in m^+$. Thus $m^+ \in S$. \subparagraph{Case 3}% Suppose $n \in m$. Then $n \in m \cup \{m\} = m^+$. Thus $m^+ \in S$. \subparagraph{Conclusion}% Since the above three cases are exhaustive, it follows $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:trichotomy-law-natural-numbers-i} and \nameref{par:trichotomy-law-natural-numbers-ii}, $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Thus for all $m, n \in \omega$, $$m \in n \lor m = n \lor n \in m.$$ We now prove that $$\in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\}$$ is \nameref{ref:irreflexive} and \nameref{ref:connected}. Irreflexivity immediately follows from \nameref{sub:lemma-4l-b}. Connectivity follows immediately from the fact $S = \omega$. Furthermore, it is not possible both $m \in n$ and $n \in m$ since, by \nameref{sub:theorem-4f}, $m$ and $n$ are \nameref{ref:transitive-set}s. This would otherwise imply $m \in m$, an immediate contradiction to irreflexivity. Thus $\in_\omega$ is a \nameref{ref:trichotomous} relation. \end{proof} \subsection{\unverified{% Linear Ordering on \texorpdfstring{$\omega$}{Natural Numbers}}}% \hyperlabel{sub:linear-ordering-natural-numbers} \begin{theorem} \nameref{ref:relation} \begin{equation} \hyperlabel{sub:linear-ordering-natural-numbers-eq1} \in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\} \end{equation} is a \nameref{ref:linear-ordering} on $\omega$. \end{theorem} \lean{Mathlib/Init/Algebra/Order} {LinearOrder.isStrictTotalOrder\_of\_linearOrder} \begin{proof} By definition, \eqref{sub:linear-ordering-natural-numbers-eq1} is a linear ordering on $\omega$ if it is (i) transitive and (ii) trichotomous. \paragraph{(i)}% Suppose $p, q, r \in \omega$ such that $p \in q$ and $q \in r$. By \nameref{sub:theorem-4f}, $p$, $q$, and $r$ are \nameref{ref:transitive-set}s. By definition of a transitive set, it follows $p \in r$. Hence \eqref{sub:linear-ordering-natural-numbers-eq1} is \nameref{ref:transitive}. \paragraph{(ii)}% By \nameref{sub:trichotomy-law-natural-numbers}, \eqref{sub:linear-ordering-natural-numbers-eq1} is trichotomous. \end{proof} \subsection{\unverified{Corollary 4M}}% \hyperlabel{sub:corollary-4m} \begin{corollary}[4M] For any natural numbers $m$ and $n$, \begin{equation} \hyperlabel{sub:corollary-4m-eq1} m \in n \iff m \subset n \end{equation} and \begin{equation} \hyperlabel{sub:corollary-4m-eq2} m \ineq n \iff m \subseteq n. \end{equation} \end{corollary} \begin{proof} \paragraph{\eqref{sub:corollary-4m-eq1}}% We prove both directions of the biconditional specified in \eqref{sub:corollary-4m-eq1}: \subparagraph{($\Rightarrow$)}% Suppose $m \in n$ and $t \in m$. By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}. Therefore $t \in n$. Hence $m \subseteq n$. Since $m \in n$, \nameref{sub:linear-ordering-natural-numbers} implies $m \neq n$. Thus, by definition of \nameref{ref:proper-subset}, $m \subset n$. \subparagraph{($\Leftarrow$)}% Suppose $m \subset n$. By \nameref{sub:linear-ordering-natural-numbers}, exactly one of $$m \in n, \quad m = n, \quad n \in m$$ holds. By definition of \nameref{ref:proper-subset}, $m \subseteq n$ and $m \neq n$. Furthermore, it cannot be that $n \in m$ since otherwise $n \in n$, contradicting \nameref{sub:lemma-4l-b}. Thus $m \in n$ is the only possibility. \paragraph{\eqref{sub:corollary-4m-eq2}}% We prove both directions of the biconditional specified in \eqref{sub:corollary-4m-eq2}: \subparagraph{($\Rightarrow$)}% Suppose $m \ineq n$. By definition, $m \in n$ or $m = n$. Let $p \in m$. Then $p \in m \in n$ or $p \in m = n$. By \nameref{sub:theorem-4f}, $n$ is a \nameref{ref:transitive-set}. Thus $p \in n$ in either case. Hence $m \subseteq n$. \subparagraph{($\Leftarrow$)}% Suppose $m \subseteq n$. By \nameref{sub:linear-ordering-natural-numbers}, exactly one of $$m \in n, \quad m = n, \quad n \in m$$ holds. But it cannot be that $n \in m$ since that would imply $n \in n$, contradicting \nameref{sub:lemma-4l-b}. Therefore $m \in n$ or $m = n$. Hence $m \ineq n$. \end{proof} \subsection{\verified{Theorem 4N}}% \hyperlabel{sub:theorem-4n} \begin{theorem}[4N] For any natural numbers $n$, $m$, and $p$, \begin{equation} m \in n \iff m + p \in n + p. \tag{i} \end{equation} If, in addition, $p \neq 0$, then \begin{equation} m \in n \iff m \cdot p \in n \cdot p. \tag{ii} \end{equation} \end{theorem} \code{Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4n\_i} \lean{Std/Data/Nat/Lemmas}{Nat.add\_lt\_add\_iff\_right} \code{Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.theorem\_4n\_ii} \lean{Init/Data/Nat/Basic}{Nat.mul\_lt\_mul\_of\_pos\_right} \begin{proof} \paragraph{(i)}% \hyperlabel{par:theorem-4n-i} Let $m$ and $n$ be \nameref{ref:natural-number}s. \subparagraph{($\Rightarrow$)}% \hyperlabel{spar:theorem-4n-i-right} Suppose $m \in n$. Let $$S = \{p \in \omega \mid m + p \in n + p\}.$$ It trivially follows that $0 \in S$. Next, suppose $p \in S$. That is, suppose $m + p \in n + p$. By \nameref{sub:lemma-4l-a}, this holds if and only if $(m + p)^+ \in (n + p)^+$. \nameref{sub:theorem-4i} then implies that $m + p^+ \in n + p^+$ meaning $p^+ \in S$. Thus $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. Therefore, for all $p \in \omega$, $m \in n$ implies $m + p \in n + p$. \subparagraph{($\Leftarrow$)}% Let $p$ be a natural number and suppose $m + p \in n + p$. By the \nameref{sub:trichotomy-law-natural-numbers}, there are two cases to consider regarding how $m$ and $n$ relate to one another: \vspace{8pt} \textbf{Case 1}: Suppose $m = n$. Then $m + p \in n + p = m + p$. \nameref{sub:lemma-4l-b} shows this is impossible. \vspace{8pt} \textbf{Case 2}: Suppose $n \in m$. Then \nameref{spar:theorem-4n-i-right} indicates $n + p \in m + p$. But this contradicts \nameref{sub:trichotomy-law-natural-numbers} since, by hypothesis, $m + p \in n + p$. \vspace{8pt} \textbf{Conclusion}: By trichotomy, it follows $m \in n$. \paragraph{(ii)}% \hyperlabel{par:theorem-4n-ii} Let $m$ and $n$ be \nameref{ref:natural-number}s. \subparagraph{($\Rightarrow$)}% \hyperlabel{spar:theorem-4n-ii-right} Suppose $m \in n$. Let $$S = \{p \in \omega \mid m \cdot p^+ \in n \cdot p^+\}.$$ $0 \in S$ by \nameref{sub:right-multiplicative-identity}. Next, suppose $p \in S$. That is, $m \cdot p^+ \in n \cdot p^+$. Then \begin{align*} m \cdot p^{++} & = m \cdot p^+ + m & \textref{sub:theorem-4j} \\ & \in n \cdot p^+ + m & \textref{par:theorem-4n-i} \\ & = m + n \cdot p^+ & \textref{sub:theorem-4k-2} \\ & \in n + n \cdot p^+ & \textref{par:theorem-4n-i} \\ & = n \cdot p^+ + n & \textref{sub:theorem-4k-2} \\ & = n \cdot p^{++}. & \textref{sub:theorem-4j} \end{align*} Therefore $p^+ \in S$. Thus $S$ is an \nameref{ref:inductive-set}. Hence \nameref{sub:theorem-4b} implies $S = \omega$. By \nameref{sub:theorem-4c}, every natural number except 0 is the successor of some natural number. Therefore, for all $p \in \omega$ such that $p \neq 0$, $m \in n$ implies $m \cdot p \in n \cdot p$. \subparagraph{($\Leftarrow$)}% Let $p \neq 0$ be a natural number and suppose $m \cdot p \in n \cdot p$. By the \nameref{sub:trichotomy-law-natural-numbers}, there are two cases to consider regarding how $m$ and $n$ relate to one another: \vspace{8pt} \textbf{Case 1}: Suppose $m = n$. Then $m \cdot p \in n \cdot p = m \cdot p$. \nameref{sub:lemma-4l-b} shows this is impossible. \vspace{8pt} \textbf{Case 2}: Suppose $n \in m$. Then \nameref{spar:theorem-4n-ii-right} indicates $n \cdot p \in m \cdot p$. But this contradicts \nameref{sub:trichotomy-law-natural-numbers} since, by hypothesis, $m \cdot p \in n \cdot p$. \vspace{8pt} \textbf{Conclusion}: By trichotomy, it follows $m \in n$. \end{proof} \subsection{\verified{Corollary 4P}}% \hyperlabel{sub:corollary-4p} \begin{corollary}[4P] The following cancellation laws hold for $m$, $n$, and $p$ in $\omega$: \begin{align} m + p = n + p & \Rightarrow m = n, & \hyperlabel{sub:corollary-4p-eq1} \\ m \cdot p = n \cdot p \land p \neq 0 & \Rightarrow m = n. & \hyperlabel{sub:corollary-4p-eq2} \end{align} \end{corollary} \code{Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.corollary\_4p\_i} \lean{Init/Data/Nat/Basic}{Nat.add\_right\_cancel} \code{Common/Nat/Basic}{Nat.mul\_right\_cancel} \begin{proof} \paragraph{\eqref{sub:corollary-4p-eq1}}% Suppose $m + p = n + p$. By the \nameref{sub:trichotomy-law-natural-numbers}, there are two cases to consider regarding how $m$ and $n$ relate to one another. If $m \in n$, then \nameref{sub:theorem-4n} implies $m + p \in n + p$. If $n \in m$, then \nameref{sub:theorem-4n} implies $n + p \in m + p$. Both of these contradict the \nameref{sub:trichotomy-law-natural-numbers} of $m + p$ and $n + p$. Thus $m = n$ is the only remaining possibility. \paragraph{\eqref{sub:corollary-4p-eq2}}% Suppose $m \cdot p = n \cdot p$ and $p \neq 0$. By the \nameref{sub:trichotomy-law-natural-numbers}, there are two cases to consider regarding how $m$ and $n$ relate to one another. If $m \in n$, then \nameref{sub:theorem-4n} implies $m \cdot p \in n \cdot p$. If $n \in m$, then \nameref{sub:theorem-4n} implies $n \cdot p \in m \cdot p$. Both of these contradict the \nameref{sub:trichotomy-law-natural-numbers} of $m \cdot p$ and $n \cdot p$. Thus $m = n$ is the only remaining possibility. \end{proof} \subsection{\verified{% Well Ordering of \texorpdfstring{$\omega$}{Natural Numbers}}}% \hyperlabel{sub:well-ordering-natural-numbers} \begin{theorem} Let $A$ be a nonempty subset of $\omega$. Then there is some $m \in A$ such that $m \ineq n$ for all $n \in A$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.well\_ordering\_nat} \lean{Mathlib/SetTheory/Ordinal/Basic}{WellOrder} \begin{note} This proof was written a few days after reading Enderton's proof as a means of ensuring I remember the main arguments. \end{note} \begin{proof} Let $A$ be a nonempty subset of $\omega$. For the sake of contradiction, suppose $A$ does not have a least element. It then suffices to prove that the complement of $A$ equals $\omega$. If we do so, then $A = \emptyset$, a contradiction. Define \begin{equation} \hyperlabel{sub:well-ordering-natural-numbers-eq1} S = \{n \in \omega \mid (\forall m \in n) m \not\in A\}. \end{equation} We prove $S$ is an \nameref{ref:inductive-set} by showing that (i) $0 \in S$ and (ii) if $n \in S$, then $n^+ \in S$. Afterward we show that $\omega - A = \omega$, completing the proof. \paragraph{(i)}% \hyperlabel{par:well-ordering-natural-numbers-i} It vacuously holds that $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:well-ordering-natural-numbers-ii} Suppose $n \in S$. We want to prove that $$\forall m, m \in n^+ \Rightarrow m \not\in A.$$ To this end, let $m \in \omega$ such that $m \in n^+$. By definition of the \nameref{ref:successor}, $m \in n$ or $m = n$. If the former, $n \in S$ implies $m \not\in A$. If the latter, it isn't possible for $n \in A$ since the \nameref{sub:trichotomy-law-natural-numbers} would otherwise imply $n$ is the least element of $A$, which is assumed to not exist. Hence $n^+ \in S$. \paragraph{Conclusion}% By \nameref{par:well-ordering-natural-numbers-i} and \nameref{par:well-ordering-natural-numbers-ii}, $S$ is an inductive set. Since $S \subseteq \omega$, \nameref{sub:theorem-4b} implies $S = \omega$. But this immediately implies $\omega = \omega - A$ meaning $A$ is the empty set. \end{proof} \subsection{\unverified{Corollary 4Q}}% \hyperlabel{sub:corollary-4q} \begin{corollary}[4Q] There is no function $f \colon \omega \rightarrow \omega$ such that $f(n^+) \in f(n)$ for every natural number $n$. \end{corollary} \begin{proof} Let $f \colon \omega \rightarrow \omega$. Then $\emptyset \subset \ran{f} \subseteq \omega$. By the \nameref{sub:well-ordering-natural-numbers}, $\ran{f}$ must have a least element. Therefore it isn't possible $f(n^+) \in f(n)$ for all $n \in \omega$. \end{proof} \subsection{\verified{% Strong Induction Principle for \texorpdfstring{$\omega$}{Natural Numbers}}}% \hyperlabel{sub:strong-induction-principle-natural-numbers} \begin{theorem} Let $A$ be a subset of $\omega$, and assume that for every $n \in \omega$, \begin{equation} \hyperlabel{sub:strong-induction-principle-natural-numbers-eq1} \text{if every number less than } n \text{ is in } A, \text{ then } n \in A. \end{equation} Then $A = \omega$. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.strong\_induction\_principle\_nat} \begin{proof} For the sake of contradiction, suppose $\omega - A$ is a nonempty set. By \nameref{sub:well-ordering-natural-numbers}, there exists a least element $m \in \omega - A$. Then every number less than $m$ is in $A$. But then \eqref{sub:strong-induction-principle-natural-numbers-eq1} implies $m \in A$, a contradiction. Thus $\omega - A$ is an empty set meaning $A = \omega$. \end{proof} \section{Exercises 4}% \hyperlabel{sec:exercises-4} \subsection{\verified{Exercise 4.1}}% \hyperlabel{sub:exercise-4.1} Show that $1 \neq 3$ i.e., that $\emptyset^+ \neq \emptyset^{+++}$. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_1} \begin{proof} By definition, \begin{align*} 1 & = \{\emptyset\} \\ 3 & = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}. \end{align*} By the \nameref{ref:extensionality-axiom}, these two sets are trivially not equal to one another. \end{proof} \subsection{\unverified{Exercise 4.2}}% \hyperlabel{sub:exercise-4.2} Show that if $a$ is a transitive set, then $a^+$ is also a transitive set. \begin{proof} Suppose $a$ is a \nameref{ref:transitive-set}. By \nameref{sub:theorem-4e}, it follows $\bigcup \left(a^+\right) = a$. By definition of a \nameref{ref:successor}, $a^+ = a \cup \{a\}$. Thus it immediately follows $$\bigcup \left(a^+\right) = a \subseteq a \cup \{a\} = a^+.$$ Therefore $a^+$ is indeed a transitive set. \end{proof} \subsection{\unverified{Exercise 4.3}}% \hyperlabel{sub:exercise-4.3} \begin{enumerate}[(a)] \item Show that if $a$ is a transitive set, then $\powerset{a}$ is also a transitive set. \item Show that if $\powerset{a}$ is a transitive set, then $a$ is also a transitive set. \end{enumerate} \begin{proof} \paragraph{(a)}% Suppose $a$ is a \nameref{ref:transitive-set}. We show that $\bigcup \powerset{a} \subseteq \powerset{a}$. Let $t \in \bigcup \powerset{a}$. By definition of the \nameref{ref:power-set}, there exists some $X \subseteq a$ such that $t \in X$. Thus $t \in a$. Because $a$ is a transitive set, every member of $t$ is a member of $a$. In other words, $t \subseteq a$. Equivalently, $t \in \powerset{a}$. \paragraph{(b)}% Suppose $\powerset{a}$ is a transitive set. We show that $\bigcup a \subseteq a$. Let $t \in \bigcup a$. Then there exists some $b \in a$ such that $t \in b$. Since $\{b\}$ is a member of $\powerset{a}$ and $\powerset{a}$ is a transitive set, $b \in \powerset{a}$. That is, $b \subseteq a$. Thus $t \in a$. \end{proof} \subsection{\unverified{Exercise 4.4}}% \hyperlabel{sub:exercise-4.4} Show that if $a$ is a transitive set, then $\bigcup a$ is also a transitive set. \begin{proof} Suppose $a$ is a transitive set. We show that $\bigcup\bigcup{a} \subseteq \bigcup a$. Let $t \in \bigcup\bigcup{a}$. Then there exists some $b \in \bigcup{a}$ such that $t \in b$. Since $a$ is transitive, $\bigcup{a} \subseteq a$. Thus $b \in a$ and $t \in \bigcup a$. \end{proof} \subsection{\unverified{Exercise 4.5}}% \hyperlabel{sub:exercise-4.5} Assume that every member of $\mathscr{A}$ is a transitive set. \begin{enumerate}[(a)] \item Show that $\bigcup\mathscr{A}$ is a transitive set. \item Show that $\bigcap\mathscr{A}$ is a transitive set (assume that $\mathscr{A}$ is nonempty). \end{enumerate} \begin{proof} \paragraph{(a)}% Suppose every member of $\mathscr{A}$ is a transitive set. We show that $\bigcup\bigcup{\mathscr{A}} \subseteq \bigcup{\mathscr{A}}$. Let $t \in \bigcup\bigcup{\mathscr{A}}$. Then there exists some $b_1 \in \bigcup{\mathscr{A}}$ such that $t \in b_1$. Likewise there exists some $b_2 \in \mathscr{A}$ such that $b_1 \in b_2$. By hypothesis, $b_2$ is transitive meaning $t \in b_2$. Thus $t \in \bigcup{\mathscr{A}}$. \paragraph{(b)}% Suppose every member of nonempty set $\mathscr{A}$ is a transitive set. We show that $\bigcup\bigcap{\mathscr{A}} \subseteq \bigcap{\mathscr{A}}$. Let $t \in \bigcup\bigcap{\mathscr{A}}$. Then there exists some $b \in \bigcap{\mathscr{A}}$ such that $t \in b$. Thus $b$ is a member of every member of $\mathscr{A}$. By hypothesis, every member of $\mathscr{A}$ is a transitive set meaning $t$ must be a member of every member of $\mathscr{A}$. In other words, $t \in \bigcap{\mathscr{A}}$. \end{proof} \subsection{\unverified{Exercise 4.6}}% \hyperlabel{sub:exercise-4.6} Prove the converse to \nameref{sub:theorem-4e}: If $\bigcup \left(a^+\right) = a$, then $a$ is a transitive set. \begin{proof} Let $a$ be a set such that $\bigcup \left(a^+\right) = a$. Then \begin{align*} \bigcup \left(a^+\right) & = \bigcup (a \cup \{a\}) & \textref{ref:successor} \\ & = \bigcup a \cup \bigcup \{a\} & \textref{sub:exercise-2.21} \\ & = a. & \text{by hypothesis} \end{align*} But $$\bigcup{a} \subseteq \bigcup a \cup \bigcup \{a\} = a.$$ Thus $a$ is indeed a \nameref{ref:transitive-set}. \end{proof} \subsection{\unverified{Exercise 4.7}}% \hyperlabel{sub:exercise-4.7} Complete part 4 of the proof of the \nameref{sub:recursion-theorem-natural-numbers}. \begin{proof} Refer to \nameref{par:recursion-theorem-natural-numbers-iv}. \end{proof} \subsection{\unverified{Exercise 4.8}}% \hyperlabel{sub:exercise-4.8} Let $f$ be a one-to-one function from $A$ into $A$, and assume that $c \in A - \ran{f}$. Define $h \colon \omega \rightarrow A$ by recursion: \begin{align*} h(0) & = c, \\ h(n^+) & = f(h(n)). \end{align*} Show that $h$ is one-to-one. \begin{proof} Let $$S = \{x \in \omega \mid \forall y, \left[ h(x) = h(y) \Rightarrow x = y \right]\}.$$ We prove that (i) $S$ is an \nameref{ref:inductive-set} and (ii) that $h$ is one-to-one. \paragraph{(i)}% \hyperlabel{par:exercise-4.8-i} We first show that $0 \in S$. Suppose there exists some $n \in \omega$ such that $h(0) = c = h(n)$. For the sake of contradiction, suppose $n \neq 0$. By \nameref{sub:theorem-4c}, there exists some $m$ such that $m^+ = n$. Then $$h(n) = h(m^+) = f(h(m)) = c.$$ But $c \in A - \ran{f}$, meaning the previous identity is an impossibility. Thus $n = 0$, i.e. $0 \in S$. Next, suppose $y, n \in \omega$ such that $n \in S$ and $h(n^+) = h(y)$. We must show $n^+ \in S$. There are two cases to consider: \subparagraph{Case 1}% Suppose $y = 0$. Then $h(y) = h(0) = c = h(n^+) = f(h(n))$. Since $c \in A - \ran{f}$, $f(h(n)) \neq c$. Thus $y \neq 0$, a contradiction. \subparagraph{Case 2}% Suppose $y \neq 0$. \nameref{sub:theorem-4c} implies there exists some $z \in \omega$ such that $z^+ = y$. Thus $$h(n^+) = f(h(n)) = f(h(z)) = h(z^+).$$ But $f$ is one-to-one meaning $h(n) = h(z)$. Since $n \in S$, $h(n) = h(z)$ implies $n = z$ which in turn implies $n^+ = z^+ = y$. Thus $n^+$ is in $S$. \paragraph{(ii)}% By \nameref{par:exercise-4.8-i}, $S \subseteq \omega$ is an inductive set. Then \nameref{sub:theorem-4b} states $S = \omega$. Hence $h$ is one-to-one. \end{proof} \subsection{\unverified{Exercise 4.9}}% \hyperlabel{sub:exercise-4.9} Let $f$ be a function from $B$ into $B$, and assume that $A \subseteq B$. We have two possible methods for constructing the "closure" $C$ of $A$ under $f$. First define $C^*$ to be the intersection of the closed supersets of $A$: \begin{equation} \hyperlabel{sub:exercise-4.9-eq1} C^* = \bigcap\{X \mid A \subseteq X \subseteq B \land \img{f}{X} \subseteq X\}. \end{equation} Alternatively, we could apply the recursion theorem to obtain the function $h$ for which \begin{align*} h(0) & = A, \\ h(n^+) & = h(n) \cup \img{f}{h(n)}. \end{align*} Clearly $h(0) \subseteq h(1) \subseteq \cdots$; define $C_*$ to be $\bigcup\ran{h}$; in other words \begin{equation} \hyperlabel{sub:exercise-4.9-eq2} C_* = \bigcup_{i \in \omega} h(i). \end{equation} Show that $C^* = C_*$. [\textit{Suggestion}: To show that $C^* \subseteq C_*$, show that $\img{f}{C_*} \subseteq C_*$. To show that $C_* \subseteq C^*$, use induction to show that $h(n) \subseteq C^*$.] \begin{proof} We show that $C^* \subseteq C_*$ and $C^* \supseteq C_*$. \paragraph{($\subseteq$)}% It suffices to show $\img{f}{C_*} \subseteq C_*$ since then $C_*$ is a member of the family of sets being intersected in \eqref{sub:exercise-4.9-eq1}. Let $t \in \img{f}{C_*}$. By definition of the \nameref{ref:image} of a set, there exists some $u \in C_*$ such that $f(u) = t$. By \eqref{sub:exercise-4.9-eq2}, there exists some $i \in \omega$ such that $u \in h(i)$. Then $t \in \img{f}{h(i)} \subseteq h(i) \cup \img{f}{h(i)} = h(i^+)$. Therefore \eqref{sub:exercise-4.9-eq2} indicates $t \in C_*$. \paragraph{($\supseteq$)}% Define $$S = \{n \in \omega \mid h(n) \subseteq C^*\}.$$ We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. Afterward we prove that (iii) $C_* \subseteq C^*$. \subparagraph{(i)}% \label{spar:exercise-4.9-i} By construction, $h(0) = A$. It trivially follows that $A \subseteq C^*$ by \eqref{sub:exercise-4.9-eq1}. Thus $h(0) \subseteq C^*$ meaning $0 \in S$. \subparagraph{(ii)}% \label{spar:exercise-4.9-ii} Suppose $n \in S$. That is, $h(n) \subseteq C^*$. We must prove that $h(n^+) \subseteq C^*$. Let $$t \in h(n^+) = h(n) \cup \img{f}{h(n)}.$$ Then either $t \in h(n)$ or $t \in \img{f}{h(n)}$. If $t \in h(n)$, it immediately follows $t \in C^*$. If $t \in \img{f}{h(n)}$, then the definition of the \nameref{ref:image} of a set implies there exists some $u \in h(n)$ such that $f(u) = t$. Since $h(n) \subseteq C^*$, \eqref{sub:exercise-4.9-eq1} indicates that $$\forall X, A \subseteq X \subseteq B \land \img{f}{X} \subseteq X \Rightarrow u \in X.$$ But then closure under image yields $$\forall X, A \subseteq X \subseteq B \land \img{f}{X} \subseteq X \Rightarrow f(u) \in X.$$ Since $f(u) = t$, $t \in C^*$. Thus $h(n^+) \subseteq C^*$, i.e. $n^+ \in S$. \subparagraph{(iii)}% By \nameref{spar:exercise-4.9-i} and \nameref{spar:exercise-4.9-ii}, $S \subseteq \omega$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. That is, for all $n \in \omega$, $h(n) \subseteq C^*$. Thus $$C_* = \bigcup_{i \in \omega} h(i) \subseteq C^*.$$ \paragraph{Conclusion}% Since $C^* \subseteq C_*$ and $C_* \subseteq C^*$, it follows $C^* = C_*$. \end{proof} \subsection{\unverified{Exercise 4.10}}% \hyperlabel{sub:exercise-4.10} In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x^2$, and $A$ is the closed interval $\icc{\frac{1}{2}}{1}$. What is the set called $C^*$ and $C_*$? \begin{proof} By \nameref{sub:exercise-4.9}, $C^* = C_*$. By definition, $$C^* = \bigcap \{X \mid \icc{\frac{1}{2}}{1} \subseteq X \subseteq \mathbb{R} \land \img{f}{X} \subseteq X\}.$$ Since $f(x)$ converges to $0$ for all values $x < 1$, it follows $C^* = C_* = \ioc{0}{1}$. \end{proof} \subsection{\unverified{Exercise 4.11}}% \hyperlabel{sub:exercise-4.11} In Exercise 9, assume that $B$ is the set of real numbers, $f(x) = x - 1$, and $A = \{0\}$. What is the set called $C^*$ and $C_*$? \begin{proof} By \nameref{sub:exercise-4.9}, $C^* = C_*$. By definition, $$C_* = \bigcup_{i \in \omega} h(i)$$ where \begin{align*} h(0) & = A = \{0\}, \\ h(n^+) & = h(n) \cup \img{f}{h(n)}. \end{align*} Thus $C_* = C^* = \{x \in \mathbb{Z} \mid x \leq 0\}$. \end{proof} \subsection{\sorry{Exercise 4.12}}% \hyperlabel{sub:exercise-4.12} Formulate an analogue to Exercise 9 for a function $f \colon B \times B \rightarrow B$. \begin{proof} TODO \end{proof} \subsection{\verified{Exercise 4.13}}% \hyperlabel{sub:exercise-4.13} Let $m$ and $n$ be natural numbers such that $m \cdot n = 0$. Show that either $m = 0$ or $n = 0$. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_13} \begin{proof} Suppose $m \cdot n = 0$. For the sake of contradiction, assume $m \neq 0$ and $n \neq 0$. By \nameref{sub:theorem-4c}, there exists some $p, q \in \omega$ such that $p^+ = m$ and $q^+ = n$. Thus \begin{align*} m \cdot n & = m \cdot q^+ \\ & = m \cdot q + m & \textref{sub:theorem-4j} \\ & = m \cdot q + p^+ \\ & = (m \cdot q + p)^+. & \textref{sub:theorem-4i} \end{align*} By definition of a \nameref{ref:successor}, $m \cdot n = (m \cdot q + p)^+ \neq \emptyset = 0$, a contradiction. Therefore our original assumption was wrong. Hence $m = 0$ or $n = 0$. \end{proof} \subsection{\verified{Exercise 4.14}}% \hyperlabel{sub:exercise-4.14} Call a natural number \textit{even} if it has the form $2 \cdot m$ for some $m$. Call it \textit{odd} if it has the form $(2 \cdot p) + 1$ for some $p$. Show that each natural number is either even or odd, but never both. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_14} \begin{proof} Let $$S = \{n \in \omega \mid n \text{ is even or odd but not both}\}.$$ We show that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$ as well. Afterward we prove (iii) that the theorem statement holds. \paragraph{(i)}% \hyperlabel{par:exercise-4.14a-i} $0$ is even since $2 \cdot 0 = 0$ by \nameref{sub:theorem-4j}. Furthermore, $0$ is not odd since that would imply there exists some $p$ such that $(2 \cdot p)^+ = 0$. By definition of \nameref{ref:successor}, this is not possible. Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:exercise-4.14a-ii} Suppose $n \in S$. Then $n$ is even or odd but not both. \subparagraph{Case 1}% Suppose $n$ is even and not odd. Then there exists some $m \in \omega$ such that $2 \cdot m = n$. Therefore $(2 \cdot m)^+ = n^+$. Hence $n^+$ is odd. For the sake of contradiction, suppose $n^+$ is even. Then there exists some $p$ such that $2 \cdot p = n^+$. We consider two additional cases: \vspace{8pt}\quad \textbf{Case 1a}: Suppose $p = 0$. Then, by \nameref{sub:theorem-4j}, $2 \cdot p = 0 = n^+$. By definition of \nameref{ref:successor}, this is not possible. \vspace{8pt}\quad \textbf{Case 1b}: Suppose $p \neq 0$. Then \nameref{sub:theorem-4c} implies there exists some $q$ such that $q^+ = p$. Thus \begin{align*} n^+ & = 2 \cdot p \\ & = 2 \cdot q^+ \\ & = q^+ + q^+ \\ & = (q^+ + q)^+. & \textref{sub:theorem-4i} \end{align*} By \nameref{sub:theorem-4d}, the \nameref{ref:successor} operation is one-to-one meaning $n = q^+ + q$. But then \begin{align*} n & = q^+ + q \\ & = q + q^+ & \textref{sub:theorem-4k-2} \\ & = (q + q)^+ & \textref{sub:theorem-4i} \\ & = (2 \cdot q)^+, \end{align*} indicating $n$ is odd. This is a contradiction. \vspace{8pt}\quad \textbf{Conclusion}: Since the above two cases are exhaustive, it follows our original assumption is wrong. That is, $n^+$ is odd but not even. \subparagraph{Case 2}% Suppose $n$ is odd and not even. Then there exists some $p \in \omega$ such that $(2 \cdot p)^+ = n$. Therefore $(2 \cdot p)^{++} = (2 \cdot p^+) = n^+$. Hence $n^+$ is even. For the sake of contradiction, suppose $n^+$ is odd. Then there exists some $q$ such that $(2 \cdot q)^+ = n^+$. By \nameref{sub:theorem-4d}, the \nameref{ref:successor} operation is one-to-one meaning $2 \cdot q = n$. But this implies $n$ is even, a contradiction. Thus our original assumption is wrong. That is, $n^+$ is even but not odd. \subparagraph{Conclusion}% Since the foregoing cases are exhaustive, it follows $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:exercise-4.14a-i} and \nameref{par:exercise-4.14a-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus every natural number is either even or odd, but not both. \end{proof} \subsection{\verified{Exercise 4.15}}% \hyperlabel{sub:exercise-4.15} Complete the proof of \nameref{sub:theorem-4k-1}. \begin{proof} Refer to \nameref{sub:theorem-4k-1}. \end{proof} \subsection{\verified{Exercise 4.16}}% \hyperlabel{sub:exercise-4.16} Complete the proof of \nameref{sub:theorem-4k-5}. \begin{proof} Refer to \nameref{sub:theorem-4k-5}. \end{proof} \subsection{\verified{Exercise 4.17}}% \hyperlabel{sub:exercise-4.17} Prove that $m^{n+p} = m^n \cdot m^p$. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_17} \lean{Data/Nat/Lemmas}{Nat.pow\_add} \begin{proof} Let $m$ and $n$ be \nameref{ref:natural-number}s and define \begin{equation} \hyperlabel{sub:exercise-4.17-eq1} S = \{p \in \omega \mid m^{n+p} = m^n \cdot m^p\}. \end{equation} We prove that (i) $0 \in S$ and (ii) if $p \in S$ then $p^+ \in S$. Afterwards we show that (iii) our theorem holds. \paragraph{(i)}% \hyperlabel{par:exercise-4.17-i} Consider $m^{n+0}$: \begin{align*} m^{n+0} & = m^n & \textref{sub:theorem-4i} \\ & = m^n \cdot 1 & \textref{sub:right-multiplicative-identity} \\ & = m^n \cdot m^0. & \textref{ref:exponentiation} \end{align*} Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:exercise-4.17-ii} Suppose $p \in S$. Now consider $m^{n+p^+}$: \begin{align*} m^{n+p^+} & = m^{(n + p)^+} & \textref{sub:theorem-4i} \\ & = E_m(n + p) \cdot m & \textref{ref:exponentiation} \\ & = m^n \cdot m^p \cdot m & \eqref{sub:exercise-4.17-eq1} \\ & = m^n \cdot (m^p \cdot m) & \textref{sub:theorem-4k-4} \\ & = m^n \cdot m^{p^+}. & \textref{ref:exponentiation} \end{align*} Thus $p^+ \in S$. \paragraph{(iii)}% By \nameref{par:exercise-4.17-i} and \nameref{par:exercise-4.17-ii}, $S \subseteq \omega$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n, p \in \omega$, it follows that $m^{n+p} = m^n \cdot m^p$. \end{proof} \subsection{\unverified{Exercise 4.18}}% \hyperlabel{sub:exercise-4.18} Simplify $\img{\in_\omega^{-1}}{\{7, 8\}}$. \begin{proof} By definition, $$\in_\omega = \{\tuple{m, n} \in \omega \times \omega \mid m \in n\}.$$ Thus, the \nameref{ref:inverse} of $\in_\omega$ is $$\in_\omega^{-1} = \{\tuple{n, m} \in \omega \times \omega \mid m \in n\}.$$ Therefore $$\img{\in_\omega^{-1}}{\{7, 8\}} = \{6, 7\}.$$ \end{proof} \subsection{\verified{Exercise 4.19}}% \hyperlabel{sub:exercise-4.19} Prove that if $m$ is a natural number and $d$ is a nonzero number, then there exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r$ is less than $d$. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_19} \begin{proof} Let $d \in \omega$ such that $d \neq 0$. Define \begin{equation} \hyperlabel{sub:exercise-4.18-eq1} S = \{m \in \omega \mid (\exists q, d \in \omega) m = (d \cdot q) + r \land r < d\}. \end{equation} We prove that $S$ is an \nameref{ref:inductive-set} by showing (i) $0 \in S$ and (ii) if $m \in S$, then $m^+ \in S$. Afterward we prove (iii) the theorem statement. \paragraph{(i)}% \hyperlabel{par:exercise-4.19-i} Let $q = 0$ and $r = 0$. We note $r < d$ by \textref{sub:zero-least-natural-number}. Furthermore, \begin{align*} (d \cdot 0) + 0 & = d \cdot 0 & \textref{sub:theorem-4i} \\ & = 0. & \textref{sub:theorem-4j} \end{align*} Thus $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:exercise-4.19-ii} Suppose $m \in S$. Then there exists $q, d \in \omega$ such that $m = (d \cdot q) + r$ and $r < d$. Then \begin{align*} m^+ & = ((d \cdot q) + r)^+ \\ & = (d \cdot q) + r^+. & \textref{sub:theorem-4i} \end{align*} By \nameref{sub:trichotomy-law-natural-numbers}, there are three cases to consider: \subparagraph{Case 1}% Suppose $r^+ \in d$. Then it immediately follows $m^+ \in S$. \subparagraph{Case 2}% Suppose $r^+ = d$. Then \begin{align*} m^+ & = (d \cdot q) + r^+ \\ & = (d \cdot q) + d \\ & = d \cdot q^+ & \textref{sub:theorem-4j} \\ & = (d \cdot q^+) + 0. & \textref{sub:theorem-4i} \end{align*} Hence $m^+ \in S$. \subparagraph{Case 3}% Suppose $d \in r^+$. Then, by definition of the \nameref{ref:successor}, $d \in r$ or $d = r$. But $r \in d$. Thus, by \nameref{sub:trichotomy-law-natural-numbers}, a contradiction is introduced. Hence this case is not possible. \subparagraph{Conclusion}% Since the above cases are exhaustive, it follows $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:exercise-4.19-i} and \nameref{par:exercise-4.19-ii}, $S \subseteq \omega$ is indeed an inductive set. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, d \in \omega$ where $d \neq 0$, there exist numbers $q$ and $r$ such that $m = (d \cdot q) + r$ and $r < d$. \end{proof} \subsection{\unverified{Exercise 4.20}}% \hyperlabel{sub:exercise-4.20} Let $A$ be a nonempty subset of $\omega$ such that $\bigcup A = A$. Show that $A = \omega$. \begin{proof} For the sake of contradiction, suppose $\omega - A$ is nonempty. By \nameref{sub:well-ordering-natural-numbers}, there exists a least number $m \in \omega$ of the set. Let $n \in \omega$ such that $n > m$. By \nameref{sub:members-natural-numbers}, $n$ is the set of all smaller natural numbers. Thus $m \in n$. Therefore $n \not\in A$ since otherwise $m \in \bigcup A$ but $m \not\in A$. Hence $\omega - A$ consists of all numbers greater than or equal to $m$. There are now two cases to consider: \paragraph{Case 1}% Suppose $m = 0$. Then $A = \emptyset$ by \nameref{sub:zero-least-natural-number}. $A$ is assumed nonempty so this is a contradiction. \paragraph{Case 2}% Suppose $m \neq 0$. Then \nameref{sub:theorem-4c} implies there exists some $p \in \omega$ such that $p^+ = m$. Because $m$ is the least element of $\omega - A$, it follows $p \in A$ is the largest element of $A$. But since $p \not\in p$ by \nameref{sub:lemma-4l-b}, $p \not\in \bigcup A$, a contradiction. \paragraph{Conclusion}% Our above two cases are exhaustive. Since both lead to contradictions, it follows our original assumption must be wrong. That is $\omega - A$ is empty, meaning $A = \omega$. \end{proof} \subsection{\unverified{Exercise 4.21}}% \hyperlabel{sub:exercise-4.21} Show that no natural number is a subset of any of its elements. \begin{proof} Let $n$ be a natural number. Suppose $m \in n$. By \nameref{sub:trichotomy-law-natural-numbers}, $n \neq m$ and $n \not\in m$. By \nameref{sub:corollary-4m}, $n \not\subseteq m$. \end{proof} \subsection{\verified{Exercise 4.22}}% \hyperlabel{sub:exercise-4.22} Show that for any natural numbers $m$ and $p$ we have $m \in m + p^+$. \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_22} \begin{proof} Let $m$ be a natural number and $$S = \{p \in \omega \mid m \in m + p^+\}.$$ We prove that (i) $0 \in S$ and (ii) if $p \in S$, then $p^+ \in S$. Afterward we prove (iii) the theorem statement. \paragraph{(i)}% \hyperlabel{par:exercise-4.22-i} By definition of the \nameref{ref:successor}, $m^+ = m \cup \{m\}$. Thus $m \in m^+ = m + 1$ (by \nameref{sub:successor-identity}). Hence $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:exercise-4.22-ii} Suppose $p \in S$. That is, $m \in m + p^+$. By definition of the \nameref{ref:successor}, \begin{align*} m + p^+ & \in (m + p^+) \cup \{m + p^+\} \\ & = (m + p^+)^+ \\ & = m + p^{++}. & \textref{sub:theorem-4i}. \end{align*} By \nameref{sub:theorem-4f}, $m + p^{++}$ is a \nameref{ref:transitive-set}. Therefore $m \in m + p^+ \in m + p^{++}$ implies $m \in m + p^{++}$. Hence $p^+ \in S$. \paragraph{(iii)}% By \nameref{par:exercise-4.22-i} and \nameref{par:exercise-4.22-ii}, $S \subseteq \omega$ is an inductive set. Thus \nameref{sub:theorem-4b} implies $S = \omega$. Hence for any natural numbers $m$ and $p$, we have $m \in m + p^+$. \end{proof} \subsection{\verified{Exercise 4.23}}% \hyperlabel{sub:exercise-4.23} Assume that $m$ and $n$ are natural numbers with $m$ less than $n$. Show that there is some $p$ in $\omega$ for which $m + p^+ = n$. (It follows from this and the preceding exercise that $m$ is less than $n$ iff $(\exists p \in \omega)m + p^+ = n$.) \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_23} \begin{proof} Let $$S = \{n \in \omega \mid (\forall m \in n, \exists p \in \omega) m + p^+ = n\}.$$ We prove that (i) $0 \in S$ and (ii) if $p \in S$, then $p^+ \in S$. Afterward we prove (iii) the theorem statement. \paragraph{(i)}% \hyperlabel{par:exercise-4.23-i} It vacuously holds that $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:exercise-4.23-ii} Suppose $n \in S$. Let $m \in n^+$. By definition of the \nameref{ref:successor}, $m \in n$ or $m = n$. If $m \in n$, then there exists some $p \in \omega$ such that $m + p^+ = n$. But then \nameref{sub:theorem-4i} implies \begin{align*} n^+ & = (m + p^+)^+ \\ & = m + p^{++}. \end{align*} If instead $m = n$, then \nameref{sub:successor-identity} implies that $$n^+ = m^+ = m + 1 = m + 0^+.$$ Hence $n^+ \in S$. \paragraph{(iii)}% By \nameref{par:exercise-4.23-i} and \nameref{par:exercise-4.23-ii}, $S$ is an \nameref{ref:inductive-set}. By \nameref{sub:theorem-4b}, $S = \omega$. Thus for all $m, n \in \omega$ such that $m \in n$, there exists some $p \in \omega$ such that $m + p^+ = n$. \end{proof} \subsection{\verified{Exercise 4.24}}% \hyperlabel{sub:exercise-4.24} Assume that $m + n = p + q$. Show that $$m \in p \iff q \in n.$$ \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_24} \begin{proof} Let $m, n, p, q \in \omega$ such that $m + n = p + q$. \paragraph{($\Rightarrow$)}% Suppose $m \in p$. By \nameref{sub:theorem-4n}, $m + n \in p + n$. By hypothesis, $m + n = p + q$ meaning $p + q \in p + n$. By \nameref{sub:theorem-4k-2}, $q + p \in n + p$. Therefore another application of \nameref{sub:theorem-4n} implies $q \in n$. \paragraph{($\Leftarrow$)}% Suppose $q \in n$. By \nameref{sub:theorem-4n}, $q + p \in n + p$. By \nameref{sub:theorem-4k-2}, $p + q \in p + n$. By hypothesis, $p + q = m + n$ meaning $m + n \in p + n$. Therefore another application of \nameref{sub:theorem-4n} implies $m \in p$. \end{proof} \subsection{\verified{Exercise 4.25}}% \hyperlabel{sub:exercise-4.25} Assume that $n \in m$ and $q \in p$. Show that $$(m \cdot q) + (n \cdot p) \in (m \cdot p) + (n \cdot q).$$ [\textit{Suggestion:} Use \nameref{sub:exercise-4.23}.] \code*{Bookshelf/Enderton/Set/Chapter\_4} {Enderton.Set.Chapter\_4.exercise\_4\_25} \begin{proof} Let $n \in m$ and $q \in p$. By \nameref{sub:exercise-4.23}, there exists some $r \in \omega$ such that $q + r^+ = p$. Then \begin{align*} & \qquad\quad n \in m \\ & \iff n \cdot r^+ \in m \cdot r^+ & \textref{sub:theorem-4n} \\ & \iff (n \cdot r^+) + ((m \cdot q) + (n \cdot q)) & \textref{sub:theorem-4n} \\ & \qquad\qquad \in (m \cdot r^+) + ((m \cdot q) + (n \cdot q)) \\ & \iff ((m \cdot q) + (n \cdot q)) + (n \cdot r^+) & \textref{sub:theorem-4k-2} \\ & \qquad\qquad \in ((m \cdot q) + (n \cdot q)) + (m \cdot r^+) \\ & \iff (m \cdot q) + ((n \cdot q) + (n \cdot r^+)) & \textref{sub:theorem-4k-1} \\ & \qquad\qquad \in ((m \cdot q) + (n \cdot q)) + (m \cdot r^+) \\ & \iff (m \cdot q) + ((n \cdot q) + (n \cdot r^+)) & \textref{sub:theorem-4k-2} \\ & \qquad\qquad \in ((n \cdot q) + (m \cdot q)) + (m \cdot r^+) \\ & \iff (m \cdot q) + ((n \cdot q) + (n \cdot r^+)) & \textref{sub:theorem-4k-1} \\ & \qquad\qquad \in (n \cdot q) + ((m \cdot q) + (m \cdot r^+)) \\ & \iff (m \cdot q) + (n \cdot (q + r^+)) & \textref{sub:theorem-4k-3} \\ & \qquad\qquad \in (n \cdot q) + (m \cdot (q + r^+)) \\ & \iff (m \cdot q) + (n \cdot p) \in (n \cdot q) + (m \cdot p) \\ & \iff (m \cdot q) + (n \cdot p) \in (m \cdot p) + (n \cdot q). & \textref{sub:theorem-4k-2} \end{align*} \end{proof} \subsection{\unverified{Exercise 4.26}}% \hyperlabel{sub:exercise-4.26} Assume that $n \in \omega$ and $f \colon n^+ \rightarrow \omega$. Show that $\ran{f}$ has a largest element. \begin{proof} By construction, the \nameref{ref:domain} of $f$ is finite. Therefore $\abs{\ran{f}} \leq \abs{\dom{f}}$, i.e. $\ran{f}$ is also a finite set. By the \nameref{sub:trichotomy-law-natural-numbers}, every member of $\ran{f}$ relates to one another. Thus there must exist a largest element. \end{proof} \subsection{\sorry{Exercise 4.27}}% \hyperlabel{sub:exercise-4.27} Assume that $A$ is a set, $G$ is a function, and $f_1$ and $f_2$ map $\omega$ into $A$. Further assume that for each $n \in \omega$ both $f_1 \restriction n$ and $f_2 \restriction n$ belong to $\dom{G}$ and $$f_1(n) = G(f_1 \restriction n) \land f_2(n) = G(f_2 \restriction n).$$ Show that $f_1 = f_2$. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 4.28}}% \hyperlabel{sub:exercise-4.28} Rewrite the proof of \nameref{sub:theorem-4g} using, in place of induction, the well-ordering of $\omega$. \begin{proof} TODO \end{proof} \setcounter{chapter}{5} \chapter{Cardinal Numbers and the Axiom of Choice}% \hyperlabel{chap:cardinal-numbers-axiom-choice} \section{Equinumerosity}% \hyperlabel{sec:equinumerosity} \subsection{\verified{Theorem 6A}}% \hyperlabel{sub:theorem-6a} \begin{theorem}[6A] For any sets $A$, $B$, and $C$, \begin{enumerate}[(a)] \item $A \equin A$. \item If $A \equin B$, then $B \equin A$. \item If $A \equin B$ and $B \equin C$, then $A \equin C$. \end{enumerate} \end{theorem} \code{Common/Set/Finite} {Set.equinumerous\_refl} \code{Common/Set/Finite} {Set.equinumerous\_symm} \code{Common/Set/Finite} {Set.equinumerous\_trans} \begin{proof} Let $A$, $B$, and $C$ be arbitrary sets. \paragraph{(a)}% Consider \nameref{ref:function} $I_A \colon A \rightarrow A$ given by $I_A(x) = x$. $I_A$ is trivially a one-to-one correspondence between $A$ and $A$. Thus $A$ is \nameref{ref:equinumerous} to $A$. \paragraph{(b)}% Suppose $A \equin B$. Then there exists a one-to-one correspondence $F$ between $A$ and $B$. Consider now \nameref{ref:inverse} $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$ By \nameref{sub:one-to-one-inverse}, $F^{-1}$ is a one-to-one function. For all $y \in A$, $\tuple{y, F(y)} \in F$. Then $\tuple{F(y), y} \in F^{-1}$ meaning $F^{-1}$ is onto $A$. Hence $F^{-1}$ is a one-to-one correspondence between $B$ and $A$, i.e. $B \equin A$. \paragraph{(c)}% Suppose $A \equin B$ and $B \equin C$. Then there exists a one-to-one correspondence $G$ between $A$ and $B$ and a one-to-one correspondence $F$ between $B$ and $C$. By \nameref{sub:one-to-one-composition}, $F \circ G$ is a one-to-one function. Thus we're left with proving $F \circ G$ is onto $C$. Let $y \in C$. Since $F$ is onto $C$, there exists some $t \in B$ such that $F(t) = y$. Likewise, since $G$ is onto $B$, there exists some $x \in A$ such that $G(x) = t$. Then $F(G(x)) = y$. Thus $\ran{(F \circ G)} = C$ meaning $F \circ G$ is onto $C$. Hence $F \circ G$ is a one-to-one correspondence function between $A$ and $C$, i.e. $A \equin C$. \end{proof} \subsection{\verified{Theorem 6B}}% \hyperlabel{sub:theorem-6b} \begin{theorem}[6B] No set is \nameref{ref:equinumerous} to its powerset. \end{theorem} \code*{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.theorem\_6b} \begin{proof} Let $A$ be an arbitrary set and $f \colon A \rightarrow \powerset{A}$. Define $\phi = \{a \in A \mid a \not\in f(a)\}$. Clearly $\phi \in \powerset{A}$. Furthermore, for all $a \in A$, $\phi \neq f(a)$ since $a \in \phi$ if and only if $a \not\in f(a)$. Thus $f$ cannot be onto $\powerset{A}$. Since $f$ was arbitrarily chosen, there exists no one-to-one correspondence between $A$ and $\powerset{A}$. Since $A$ was arbitrarily chosen, there is no set equinumerous to its powerset. \end{proof} \section{Finite Sets}% \hyperlabel{sec:finite-sets} \subsection{\verified{Pigeonhole Principle}}% \hyperlabel{sub:pigeonhole-principle} \begin{theorem} No natural number is equinumerous to a proper subset of itself. \end{theorem} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.pigeonhole\_principle} \lean{Mathlib/Data/Finset/Card} {Finset.exists\_ne\_map\_eq\_of\_card\_lt\_of\_maps\_to} \begin{proof} Let \begin{equation} \hyperlabel{sub:pigeonhole-principle-eq1} S = \{n \in \omega \mid \forall M \subset 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 definition, $0 = \emptyset$. Then $0$ has no proper subsets. Hence $0 \in S$ vacuously. \paragraph{(ii)}% \hyperlabel{par:pigeonhole-principle-ii} Suppose $n \in S$ and $M \subset n^+$. Furthermore, let $f \colon M \rightarrow n^+$ be a one-to-one \nameref{ref:function}. If $M = \emptyset$, it vacuously holds that $f$ is not onto $n^+$. Otherwise $M \neq \emptyset$. Because $M$ is finite, the \nameref{sub:trichotomy-law-natural-numbers} implies the existence of a largest member $p \in M$. There are two cases to consider: \subparagraph{Case 1}% Suppose $n \not\in \ran{f}$. Then $f$ is not onto $n^+$. \subparagraph{Case 2}% Suppose $n \in \ran{f}$. Then there exists some $t \in M$ such that $\tuple{t, n} \in f$. Define $f' \colon M \rightarrow n^+$ given by \begin{align*} f'(p) & = f(t) = n \\ f'(t) & = f(p) \\ f'(x) & = f(x) & \text{for all other } x. \end{align*} That is, $f'$ is a variant of $f$ in which the largest element of its domain (i.e. $p$) corresponds to value $n$. Next define $g = f' - \{\tuple{p, n}\}$. Then $g$ is a function mapping $M - \{p\}$ to $n$. Since $f$ is one-to-one, $f'$ and $g$ are also one-to-one. Then \eqref{sub:pigeonhole-principle-eq1} indicates $g$ must not be onto $n$. That is, there exists some $a \in n$ such that $a \not\in \ran{g}$. By the \nameref{sub:trichotomy-law-natural-numbers}, $a \neq n$. Therefore $a \not\in \ran{f'}$. $\ran{f'} = \ran{f}$ meaning $a \not\in \ran{f}$. Because $a \in n \in n^+$, \nameref{sub:theorem-4f} implies $a \in n^+$. Hence $f$ 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{\verified{Corollary 6C}}% \hyperlabel{sub:corollary-6c} \begin{corollary}[6C] No finite set is equinumerous to a proper subset of itself. \end{corollary} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.corollary\_6c} \begin{proof} Let $S$ be a \nameref{ref:finite-set} and $S'$ be a \nameref{ref:proper-subset} of $S$. Then there exists some set $T$, disjoint from $S'$, such that $S' \cup T = S$. By definition of a \nameref{ref:finite-set}, $S$ is \nameref{ref:equinumerous} to a natural number $n$. By \nameref{sub:theorem-6a}, $S' \cup T \equin S$ which, by the same theorem, implies $S' \cup T \equin n$. Let $f$ be a one-to-one correspondence between $S' \cup T$ and $n$. Then $f \restriction S'$ 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{\verified{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. \end{enumerate} \end{corollary} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.corollary\_6d\_a} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.corollary\_6d\_b} \begin{proof} \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{\verified{Corollary 6E}}% \hyperlabel{sub:corollary-6e} \begin{corollary}[6E] Any finite set is equinumerous to a unique natural number. \end{corollary} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.corollary\_6e} \begin{proof} 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 $m \equin S$ and $S \equin n$. By \nameref{sub:theorem-6a}, it follows $m \equin n$. 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{\verified{Lemma 6F}}% \hyperlabel{sub:lemma-6f} \begin{lemma}[6F] If $C$ is a proper subset of a natural number $n$, then $C \equin m$ for some $m$ less than $n$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.lemma\_6f} \begin{proof} Let \begin{equation} \hyperlabel{sub:lemma-6f-eq1} S = \{n \in \omega \mid \forall C \subset n, \exists m < n \text{ such that } C \equin 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 $C \equin n$. \end{proof} \subsection{\verified{Corollary 6G}}% \hyperlabel{sub:corollary-6g} \begin{corollary}[6G] Any subset of a finite set is finite. \end{corollary} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.corollary\_6g} \begin{proof} 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 a 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 $S' \equin m$. Hence $S'$ is a finite set. \end{proof} \subsection{\verified{Subset Size}}% \hyperlabel{sub:subset-size} \begin{lemma} Let $A$ be a finite set and $B \subseteq A$. Then there exist natural numbers $m, n \in \omega$ such that $B \equin m$, $A \equin n$, and $m \leq n$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.subset\_size} \begin{proof} Let $A$ be a \nameref{ref:finite-set} and $B$ be a subset of $A$. By \nameref{sub:corollary-6g}, $B$ must be finite. By definition of a finite set, there exists natural numbers $m, n \in \omega$ such that $B \equin m$ and $A \equin n$. By \nameref{sub:trichotomy-law-natural-numbers}, it suffices to prove that $m > n$ is not possible for then either $m < n$ or $m = n$. For the sake of contradiction, assume $m > n$. By definition of \nameref{ref:equinumerous}, there exists a one-to-one correspondence between $B$ and $m$. \nameref{sub:theorem-6a} indicates there then exists a one-to-one correspondence $f$ between $m$ and $B$. Likewise, there exists a one-to-one correspondence $g$ between $A$ and $n$. Define $h \colon A \rightarrow B$ as $h(x) = f(g(x))$ for all $x \in A$. Since $n \subset m$ by \nameref{sub:corollary-4m}, $h$ is well-defined. By \nameref{sub:one-to-one-composition}, $h$ must be one-to-one. Thus $h$ is a one-to-one correspondence between $A$ and $\ran{h}$, i.e. $A \equin \ran{h}$. But $n < m$ meaning $\ran{h} \subset B$ which in turn is a proper subset of $A$ by hypothesis. \nameref{sub:corollary-6c} states no finite set is equinumerous to a proper subset of itself, a contradiction. \end{proof} \subsection{\pending{Finite Domain and Range Size}}% \hyperlabel{sub:finite-domain-range-size} \begin{lemma} Let $A$ and $B$ be finite sets and $f \colon A \rightarrow B$ be a function. Then there exist natural numbers $m, n \in \omega$ such that $\dom{f} \equin m$, $\ran{f} \equin n$, and $m \geq n$. \end{lemma} \begin{note} This proof avoids the \nameref{ref:axiom-of-choice-1} because $A$ and $B$ are finite. In particular, we are able to choose a "smallest" element of each preimage set. Contrast this to \nameref{sub:theorem-3j-b}. \end{note} \begin{proof} Let $A$ and $B$ be \nameref{ref:finite-set}s and $f \colon A \rightarrow B$ be a function. By definition of finite sets, there exists \nameref{ref:natural-number}s $m, p \in \omega$ such that $A \equin m$ and $B \equin p$. By definition of the \nameref{ref:domain} of a function, $\dom{f} = A$. Thus $\dom{f} \equin m$. By \nameref{sub:theorem-6a}, there exists a one-to-one correspondence $g$ between $m$ and $\dom{f} = A$. For all $y \in \ran{f}$, consider $\img{f^{-1}}{\{y\}}$. Let $$A_y = \{ x \in m \mid f(g(x)) = y \}.$$ Since $g$ is a one-to-one correspondence, it follows that $A_y \equin \img{f^{-1}}{\{y\}}$. Since $A_y$ is a nonempty subset of natural numbers, the \nameref{sub:well-ordering-natural-numbers} implies there exists a least element, say $q_y$. Define $C = \{q_y \mid y \in \ran{f}\}$. Thus $h \colon C \rightarrow \ran{f}$ given by $h(x) = f(g(x))$ is a one-to-one correspondence between $C$ and $\ran{f}$ by construction. That is, $C \equin \ran{f}$. By \nameref{sub:lemma-6f}, there exists some $n \leq m$ such that $C \equin n$. By \nameref{sub:theorem-6a}, $n \equin \ran{f}$. \end{proof} \subsection{\verified{Set Difference Size}}% \hyperlabel{sub:set-difference-size} \begin{lemma} Let $A \equin m$ for some natural number $m$ and $B \subseteq A$. Then there exists some $n \in \omega$ such that $n \leq m$, $B \equin n$, and $A - B \equin m - n$. \end{lemma} \code{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.sdiff\_size} \begin{proof} Let \begin{equation} \hyperlabel{sub:set-difference-size-ih} S = \{m \in \omega \mid \forall A \equin m, \forall B \subseteq A, \exists n \in \omega (n \leq m \land B \equin n \land A - B \equin m - n)\}. \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:set-difference-size-i} Let $A \equin 0$ and $B \subseteq A$. Then it follows $A = B = \emptyset = 0$. Since $0 \leq 0$, $B \equin 0$, and $A - B = \emptyset \equin 0 = 0 - 0$, it follows $0 \in S$. \paragraph{(ii)}% \hyperlabel{par:set-difference-size-ii} Suppose $m \in S$ and consider $m^+$. Let $A \equin m^+$ and let $B \subseteq A$. By definition of \nameref{ref:equinumerous}, there exists a one-to-one correspondence $f$ between $A$ and $m^+$. Since $f$ is one-to-one and onto, there exists a unique value $a \in A$ such that $f(a) = m$. Then $B - \{a\} \subseteq A - \{a\}$ and $f$ is a one-to-one correspondence between $A - \{a\}$ and $m$. By \ihref{sub:set-difference-size-ih}, there exists some $n \in \omega$ such that $n \leq m$, $B - \{a\} \equin n$ and \begin{equation} \hyperlabel{par:set-difference-size-ii-eq1} (A - \{a\}) - (B - \{a\}) \equin m - n. \end{equation} There are two cases to consider: \subparagraph{Case 1}% Assume $a \in B$. Then $B \equin n^+$. Furthermore, by definition of the set difference, \begin{align} (A - \{a\}) & - (B - \{a\}) \nonumber \\ & = \{x \mid x \in A - \{a\} \land x \not\in B - \{a\}\} \nonumber \\ & = \{x \mid (x \in A \land x \neq a) \land \neg(x \in B \land x \neq a)\} \nonumber \\ & = \{x \mid (x \in A \land x \neq a) \land (x \not\in B \lor x = a)\} \nonumber \\ & = \{x \mid (x \in A \land x \neq a \land x \not\in B) \lor (x \in A \land x \neq a \land x = a)\} \nonumber \\ & = \{x \mid (x \in A \land x \neq a \land x \not\in B) \lor F\} \nonumber \\ & = \{x \mid (x \in A \land x \neq a \land x \not\in B)\} \nonumber \\ & = \{x \mid x \in A - B \land x \neq a\} \nonumber \\ & = \{x \mid x \in A - B \land x \not\in \{a\}\} \nonumber \\ & = (A - B) - \{a\}. \hyperlabel{par:set-difference-size-ii-eq2} \end{align} Since $a \in A$ and $a \in B$, $(A - B) - \{a\} = A - B$. Thus \begin{align*} (A - \{a\} - (B - \{a\}) & = (A - B) - \{a\} & \eqref{par:set-difference-size-ii-eq2} \\ & = A - B \\ & \equin m - n & \eqref{par:set-difference-size-ii-eq1} \\ & \equin m^+ - n^+. \end{align*} \subparagraph{Case 2}% Assume $a \not\in B$. Then $B - \{a\} = B$ (i.e. $B \approx n$) and \begin{align*} (A - \{a\}) - (B - \{a\}) & = (A - \{a\}) - B \\ & \equin m - n. & \eqref{par:set-difference-size-ii-eq1} \end{align*} The above implies that there exists a one-to-one correspondence $g$ between $(A - \{a\}) - B$ and $m - n$. Therefore $g \cup \{\tuple{a, m}\}$ is a one-to-one correspondence between $A - B$ and $(m - n) \cup \{m\}$. Hence $$A - B \equin (m - n) \cup \{m\} \equin m^+ - n.$$ \subparagraph{Subconclusion}% The above two cases are exhaustive and both conclude the existence of some $n \in \omega$ such that $n \leq m^+$, $B \equin n$ and $A - B \equin m^+ - n$. Hence $m^+ \in S$. \paragraph{(iii)}% By \nameref{par:set-difference-size-i} and \nameref{par:set-difference-size-ii}, $S \subseteq \omega$ is an \nameref{ref:inductive-set}. Thus \nameref{sub:theorem-4b} implies $S = \omega$. Hence, for all $A \equin m$ for some $m \in \omega$, if $B \subseteq A$, then there exists some $n \in \omega$ such that $n \leq m$, $B \equin n$, and $A - B \equin m - n$. \end{proof} \subsection{\sorry{Theorem 6H}}% \hyperlabel{sub:theorem-6h} Assume that $K_1 \equin K_2$ and $L_1 \equin L_2$. \begin{enumerate}[(a)] \item If $K_1 \cap L_1 = K_2 \cap L_2 = \emptyset$, then $K_1 \cup L_1 \equin K_2 \cup L_2$. \item $K_1 \times L_1 \equin K_2 \times L_2$. \item $^{(L_1)}{K_1} \equin ^{(L_2)}{K_2}$. \end{enumerate} \begin{proof} TODO \end{proof} \subsection{\sorry{Theorem 6I}}% \hyperlabel{sub:theorem-6i} For any cardinal numbers $\kappa$, $\lambda$, and $\mu$: \begin{enumerate} \item $\kappa + \lambda = \lambda + \kappa$ and $\kappa \cdot \lambda = \lambda \cdot \kappa$. \item $\kappa + (\lambda + \mu) = (\kappa + \lambda) + \mu$ and $\kappa \cdot (\lambda \cdot \mu) = (\kappa \cdot \lambda) \cdot \mu$. \item $\kappa \cdot (\lambda + \mu) = \kappa \cdot \lambda + \kappa \cdot \mu$. \item $\kappa^{\lambda + \mu} = \kappa^\lambda \cdot \kappa^\mu$. \item $(\kappa \cdot \lambda)^\mu = \kappa^\mu \cdot \lambda^\mu$. \item $(\kappa^\lambda)^\mu = \kappa^{\lambda \cdot \mu}$. \end{enumerate} \begin{proof} TODO \end{proof} \subsection{\sorry{Theorem 6J}}% \hyperlabel{sub:theorem-6j} Let $m$ and $n$ be finite cardinals. Then \begin{align*} m + n & = m +_\omega n, \\ m \cdot n = m \cdot_\omega n, \\ m^n = m^n, \end{align*} where on the right side we use the operations of $\omega$ defined via recursion and on the left side we use the operations of cardinal arithmetic. \begin{proof} TODO \end{proof} \subsection{\sorry{Corollary 6K}}% \hyperlabel{sub:corollary-6k} If $A$ and $B$ are finite, then $A \cup B$, $A \times B$, and $^B{A}$ are also finite. \begin{proof} TODO \end{proof} \section{Exercises 6}% \hyperlabel{sec:exercises-6} \subsection{\unverified{Exercise 6.1}}% \hyperlabel{sub:exercise-6.1} Show that the equation $$f(m, n) = 2^m(2n + 1) - 1$$ defines a one-one-one correspondence between $\omega \times \omega$ and $\omega$. \begin{proof} We prove that (i) $f$ is one-to-one and (ii) $f$ is onto $\omega$. \paragraph{(i)}% Let $y \in \ran{f}$. Then there exists $(m_1, n_1) \in \omega \times \omega$ such that $f(m_1, n_1) = y$. Suppose there exists $(m_2, n_2) \in \omega \times \omega$ such that $f(m_2, n_2) = y$. We show that $m_1 = m_2$ and $n_1 = n_2$. By \nameref{sub:trichotomy-law-natural-numbers}, we know either $m_1 \leq m_2$ or $m_2 \leq m_1$ and either $n_1 \leq n_2$ or $n_2 \leq n_1$. WLOG, assume $m_1 \leq m_2$ and $n_1 \leq n_2$. Then by \nameref{sub:exercise-4.23}, there exists some $p \in \omega$ such that $m_1 + p = m_2$. Likewise, there exists some $q \in \omega$ such that $n_1 + q = n_2$. Therefore \begin{align*} & f(m_1, n_1) = f(m_2, n_2) \\ & \iff 2^{m_1}(2n_1 + 1) - 1 = 2^{m_2}(2n_2 + 1) - 1 \\ & \iff 2^{m_1}(2n_1 + 1) = 2^{m_2}(2n_2 + 1) \\ & \iff 2^{m_1}(2n_1 + 1) = 2^{m_1+p}(2(n_1 + q) + 1) \\ & \iff 2^{m_1}(2n_1 + 1) = 2^{m_1}2^p(2n_1 + 2q + 1) \\ & \iff 2n_1 + 1 = 2^p(2n_1 + 2q + 1) \\ & \iff 2n_1 + 1 = 2^{p+1}n_1 + 2^{p+1}q + 2^p. \end{align*} Notice though that the right-hand side of the above equality is smallest when $p = 0$ and $q = 0$. Since when evaluated at $p = 0$ and $q = 0$, the equality holds, it follows this value of $p$ and $q$ are the only possible values that $p$ and $q$ can take on. Thus $m_1 + p = m_1 = m_2$ and $n_1 + q = n_1 = n_2$. \paragraph{(ii)}% We show $\ran{f} = \omega$ by the \nameref{sub:strong-induction-principle-natural-numbers}. It is clear that $\ran{f}$ is a subset of $\omega$. Let $x \in \omega$ and suppose every number less than $x$ is in $\ran{f}$. All that remains to be shown is that $x \in \ran{f}$. By \nameref{sub:exercise-4.14}, there are two cases to consider: \subparagraph{Case 1}% Suppose $x$ is an even number. Then $x = 2 \cdot y$ for some $y \in \omega$. Then $f(0, y) = 2^0(2y + 1) - 1 = 2y = x$. Hence $x \in \ran{f}$. \subparagraph{Case 2}% Suppose $x$ is an odd number. Then $x = (2 \cdot y) + 1$ for some $y$. It immediately follows $x - 1$ is an even number, meaning there exists some $z$ such that $x - 1 = 2z$. Since $z \in x$, the induction hypothesis states that there exists some $m, n \in \omega$ such that $$f(m, n) = z = 2^m(2n + 1) - 1.$$ Therefore \begin{align*} f(m + 1, n) & = 2^{m+1}(2n + 1) - 1 \\ & = (2)(2^m)(2n + 1) - 1 \\ & = 2z + 1 \\ & = x. \end{align*} Hence $x \in \ran{f}$. \end{proof} \subsection{\unverified{Exercise 6.2}}% \hyperlabel{sub:exercise-6.2} Show that in Fig. 32 we have: \begin{align*} J(m, n) & = [1 + 2 + \cdots + (m + n)] + m \\ & = \frac{1}{2}[(m + n)^2 + 3m + n]. \end{align*} \begin{proof} \begin{figure}[ht] \label{sub:exercise-6-2-fig1} \includegraphics[width=0.6\textwidth]{fig-32} \centering \end{figure} Let $m, n \in \omega$. We note the next point following $(m, n)$ that coincides with the $x$-axis is $(m + n, 0)$. We can then formulate the sum of points seen as \begin{equation} \hyperlabel{sub:exercise-5-2-eq1} \left[ \sum_{i=0}^{m + n} (i + 1) \right] - n. \end{equation} All that remains is showing \eqref{sub:exercise-5-2-eq1} identifies with the equation for $J$. \eqref{sub:exercise-5-2-eq1} is an arithmetic series. By \nameref{S:sub:sum-arithmetic-series}, \begin{align*} \left[ \sum_{i=0}^{m + n} (i + 1) \right] - n & = \frac{[(m + n) + 1][1 + (m + n + 1)]}{2} - n \\ & = \frac{[(m + n) + 1][(m + n) + 2]}{2} - n \\ & = \frac{(m + n)^2 + 2(m + n) + (m + n) - 2n}{2} \\ & = \frac{1}{2}[(m + n)^2 + 3m + n] \\ & = J(m, n). \end{align*} Hence $J$ is correctly defined. \end{proof} \subsection{\unverified{Exercise 6.3}}% \hyperlabel{sub:exercise-6.3} Find a one-to-one correspondence between the open unit interval $\ioo{0}{1}$ and $\mathbb{R}$ that takes rationals to rationals and irrationals to irrationals. \begin{proof} \begin{figure}[ht] \label{sub:exercise-6-3-fig1} \includegraphics[width=0.6\textwidth]{exercise-6.3.png} \centering \end{figure} Consider function $f \colon (0, 1) \rightarrow \mathbb{R}$ given by $$f(x) = \begin{cases} \frac{1}{2x} - 1 & \text{if } x \leq \frac{1}{2} \\ \frac{1}{2(x - 1)} + 1 & \text{otherwise}. \end{cases}$$ We prove that (i) $f$ is a one-to-one into $\mathbb{R}$, (ii) $f$ is onto $\mathbb{R}$, and (iii) $f$ takes rationals to rationals and irrationals to irrationals. \paragraph{(i)}% Before proceeding, consider the solutions to the following identity: \begin{align*} \frac{1}{2x} - 1 = \frac{1}{2(x - 1)} + 1 & \iff -8x^2 + 8x - 2 = 0 \\ & \iff 4x^2 - 4x + 1 = 0. \end{align*} Applying the quadratic equation shows $x = 1 / 2$ is the only solution. Thus for any $x_1, x_2 \in \ioo{0}{1}$ such that $f(x_1) = f(x_2)$, it must be that $x_1, x_2 \leq 1 / 2$ or $x_1, x_2 > 1 / 2$. We now prove $f$ is one-to-one. Let $y \in \ran{f}$. Then there exists some $x_1 \in \ioo{0}{1}$ such that $f(x_1) = y$. Suppose there exists some $x_2 \in \ioo{0}{1}$ such that $f(x_1) = f(x_2)$. We prove that $x_1 = x_2$ by case analysis: \subparagraph{Case 1}% Suppose $x_1, x_2 \leq 1 / 2$. Then $$\frac{1}{2x_1} - 1 = \frac{1}{2x_2} - 1$$ which straightforwardly simplifies to $x_1 = x_2$. \subparagraph{Case 2}% Suppose $x_1, x_2 > 1 / 2$. Then $$\frac{1}{2(x_1 - 1)} + 1 = \frac{1}{2(x_2 - 1)} + 1$$ also straightfowardly simplies to $x_1 = x_2$. \subparagraph{Subconclusion}% The above cases are exhaustive. Therefore $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Hence $f$ is one-to-one. \paragraph{(ii)}% Let $y \in \mathbb{R}$. We prove that there exists an $x \in (0, 1)$ such that $f(x) = y$. There are three cases we consider: \subparagraph{Case 1}% Suppose $y = 0$. Then $f(1 / 2) = 0 = y$ is a readily identifiable solution. \subparagraph{Case 2}% Suppose $y > 0$. Consider $x = \frac{1}{2(y + 1)}$. We note that $x < 1 / 2$ meaning \begin{align*} f(x) & = f\left(\frac{1}{2(y + 1)}\right) \\ & = \frac{1}{2(\frac{1}{2(y + 1)})} - 1 \\ & = (y + 1) - 1 \\ & = y. \end{align*} \subparagraph{Case 3}% Suppose $y < 0$. Consider $x = \frac{1}{2(y - 1)} + 1$. We note that $x > 1 / 2$ meaning \begin{align*} f(x) & = f\left(\frac{1}{2(y - 1)} + 1\right) \\ & = \frac{1}{2((\frac{1}{2(y - 1)} + 1) - 1)} + 1 \\ & = (y - 1) + 1 \\ & = y. \end{align*} \subparagraph{Subconclusion}% The above three cases are exhaustive. Thus $\ran{f} = \mathbb{R}$, i.e. $f$ is onto $\mathbb{R}$. \paragraph{(iii)}% Let $x \in (0, 1)$. There are two cases to consider: \subparagraph{Case 1}% Suppose $x$ is a rational number. Then there exist integers $m$ and $n$ such that $x = m / n$. If $x \leq 1 / 2$, then \begin{align*} f(x) & = f(m / n) \\ & = \frac{1}{2\left(\frac{m}{n}\right)} - 1 \\ & = \frac{n}{2m} - 1 \\ & = \frac{n - 2m}{2m}. \end{align*} Since $n - 2m$ and $2m$ are integers, $f(x)$ is a rational number. If instead $x > 1 / 2$, then \begin{align*} f(x) & = f(m / n) \\ & = \frac{1}{2\left(\frac{m}{n} - 1\right)} + 1 \\ & = \frac{n}{2(m - n)} + 1 \\ & = \frac{n + 2(m - n)}{2(m - n)}. \end{align*} Since $n + 2(m - n)$ and $2(m - n)$ are integers, $f(x)$ is again a rational number. Thus $f$ maps every rational number to a rational number. \subparagraph{Case 2}% Suppose $x$ is an irrational number. First, consider the case where $x \leq 1 / 2$ and, for the sake of contradiction, suppose $f(x)$ was a rational number. Then there exist integers $m$ and $n$ such that $$f(x) = \frac{1}{2x} - 1 = \frac{m}{n}.$$ But this would imply $$x = \frac{n}{2(m + n)},$$ a contradiction. Thus $f(x)$ must be irrational. Likewise, consider the case where $x > 1 / 2$. Again, for the sake of contradiction, suppose $f(x)$ was a rational number. Then there exist integers $m$ and $n$ such that $$f(x) = \frac{1}{2(x - 1)} + 1 = \frac{m}{n}.$$ But this would imply $$x = \frac{n}{2(m - n)} + 1,$$ a contradiction. Thus $f(x)$ must be irrational. Hence $f$ maps every irrational number to an irrational number. \end{proof} \subsection{\sorry{Exercise 6.4}}% \hyperlabel{sub:exercise-6.4} Construct a one-to-one correspondence between the closed unit interval $$\icc{0}{1} = \{x \in \mathbb{R} \mid 0 \leq x \leq 1\}$$ and the open unit interval $\ioo{0}{1}$. \begin{proof} TODO \end{proof} \subsection{\verified{Exercise 6.5}}% \hyperlabel{sub:exercise-6.5} Prove \nameref{sub:theorem-6a}. \begin{proof} Refer to \nameref{sub:theorem-6a}. \end{proof} \subsection{\unverified{Exercise 6.6}}% \hyperlabel{sub:exercise-6.6} Let $\kappa$ be a nonzero cardinal number. Show there does not exist a set to which every set of cardinality $\kappa$ belongs. \begin{proof} Let $\kappa$ be a nonzero cardinal number and define $$\mathbf{K}_\kappa = \{ X \mid \card{X} = \kappa \}.$$ For the sake of contradiction, suppose $\mathbf{K}_\kappa$ is a set. Then the \nameref{ref:union-axiom} suggests $\bigcup \mathbf{K}_{\kappa}$ is a set. But this "set" is precisely the class of all sets, which is \textit{not} a set. Thus our original assumption was incorrect. That is, there does not exist a set to which every set of cardinality $\kappa$ belongs. \end{proof} \subsection{\pending{Exercise 6.7}}% \hyperlabel{sub:exercise-6.7} Assume that $A$ is finite and $f \colon A \rightarrow A$. Show that $f$ is one-to-one iff $\ran{f} = A$. \code*{Bookshelf/Enderton/Set/Chapter\_6} {Enderton.Set.Chapter\_6.exercise\_6\_7} \begin{proof} Let $A$ be a \nameref{ref:finite-set} and $f \colon A \rightarrow A$. \paragraph{($\Rightarrow$)}% Suppose $f$ is one-to-one. Then $f$ is a one-to-one correspondence between $A$ and $\ran{f}$. That is, $A \equin \ran{f}$. Because $f$ maps $A$ onto $A$, $\ran{f} \subseteq A$. Hence $\ran{f} \subset A$ or $\ran{f} = A$. But, by \nameref{sub:corollary-6c}, $\ran{f}$ cannot be a proper subset of $A$. Thus $\ran{f} = A$. \paragraph{($\Leftarrow$)}% If $A = \emptyset$, then $f$ is trivially one-to-one. Assume now $A \neq \emptyset$ and $\ran{f} = A$. Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2) = y$ for some $y \in A$. We must prove that $x_1 = x_2$. Let $B = \img{f^{-1}}{\{y\}}$. Then $x_1, x_2 \in B$. Since $B \subseteq A$, \nameref{sub:subset-size} indicates that there exist natural numbers $m_1, n_1 \in \omega$ such that $B \equin m_1$, $A \equin n_1$, and $m_1 \leq n_1$. Define $f' \colon (A - B) \rightarrow (A - \{y\})$ as the \nameref{ref:restriction} of $f$ to $A - B$. Since $\ran{f} = A$, it follows $\ran{f'} = A - \{y\}$. Since \nameref{sub:corollary-6g} implies $A - B$ and $A - \{y\}$ are finite sets, \nameref{sub:finite-domain-range-size} implies the existence of natural numbers $m_2, n_2 \in \omega$ such that $\dom{f'} \equin m_2$, $\ran{f'} \equin n_2$, and $m_2 \geq n_2$. Thus, by \nameref{sub:set-difference-size}, \begin{align*} m_2 & \equin \dom{f'} \equin A - B \equin n_1 - m_1 \\ n_2 & \equin \ran{f'} \equin A - \{y\} \equin n_1 - 1. \end{align*} By \nameref{sub:corollary-6e}, $m_2 = n_1 - m_1$ and $n_2 = n_1 - 1$. Since $m_2 \geq n_2$, $n_1 - m_1 \geq n_1 - 1$. But $1 \leq m_1 \leq n_1$ meaning $m_1 = 1$. Hence $B$ is a singleton. Therefore $x_1 = x_2$, i.e. $f$ is one-to-one. \end{proof} \subsection{\pending{Exercise 6.8}}% \hyperlabel{sub:exercise-6.8} Prove that the union of two finite sets is finite, without any use of arithmetic. \begin{proof} TODO \end{proof} \subsection{\pending{Exercise 6.9}}% \hyperlabel{sub:exercise-6.9} Prove that the Cartesian product of two finite sets is finite, without any use of arithmetic. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 6.10}}% \hyperlabel{sub:exercise-6.10} Prove part 4 of \nameref{sub:theorem-6i}. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 6.11}}% \hyperlabel{sub:exercise-6.11} Prove part 5 of \nameref{sub:theorem-6i}. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 6.12}}% \hyperlabel{sub:exercise-6.12} The proof to \nameref{sub:theorem-6i} involves eight instances of showing two sets to be equinumerous. (The eight are listed in the proof of the theorem as statements numbered 1-6.) In which of these eight cases does equality actually hold? \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 6.13}}% \hyperlabel{sub:exercise-6.13} Show that a finite union of finite sets is finite. That is, show that if $B$ is a finite set whose members are themselves finite sets, then $\bigcup{B}$ is finite. \begin{proof} TODO \end{proof} \subsection{\sorry{Exercise 6.14}}% \hyperlabel{sub:exercise-6.14} Define a \textit{permutation} of $K$ to be any one-to-one function from $K$ onto $K$. We can the define the factorial operation on cardinal numbers by the equation $$\kappa! = \card{\{f \mid f \text{ is a permutation of } K\}},$$ where $K$ is any set of cardinality $\kappa$. Show that $\kappa!$ is well defined, i.e. the value of $\kappa!$ is independent of just which set $K$ is chosen. \begin{proof} TODO \end{proof} \end{document}