Enderton. Lean references of definitions and more exercises.

finite-set-exercises
Joshua Potter 2023-07-03 14:58:04 -06:00
parent 140195a8ee
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2 changed files with 119 additions and 16 deletions

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@ -20,25 +20,43 @@
\chapter{Reference}% \chapter{Reference}%
\label{chap:reference} \label{chap:reference}
\section{\pending{Axiom of Choice, First Form}}% \section{\defined{Axiom of Choice, First Form}}%
\label{ref:axiom-of-choice-1} \label{ref:axiom-of-choice-1}
For any relation $R$ there is a function $H \subseteq R$ with For any relation $R$ there is a function $H \subseteq R$ with
$\dom{H} = \dom{R}$. $\dom{H} = \dom{R}$.
\section{\pending{Axiom of Choice, Second Form}}% \begin{axiom}
\lean*{Init/Prelude}{Classical.choice}
\end{axiom}
\section{\defined{Axiom of Choice, Second Form}}%
\label{ref:axiom-of-choice-2} \label{ref:axiom-of-choice-2}
For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$ For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$ for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$
\section{\pending{Compatible}}% \begin{axiom}
\lean*{Init/Prelude}{Classical.choice}
\end{axiom}
\section{\defined{Compatible}}%
\label{sec:compatible} \label{sec:compatible}
A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and
only if for all $x$ and $y$ in $A$, only if for all $x$ and $y$ in $A$,
$$xRy \Rightarrow F(x)RF(y).$$ $$xRy \Rightarrow F(x)RF(y).$$
\begin{definition}
\lean*{Init/Core}{Quotient.lift}
\end{definition}
\section{\defined{Composition}}% \section{\defined{Composition}}%
\label{ref:composition} \label{ref:composition}
@ -75,13 +93,19 @@ There is a set having no members:
\end{axiom} \end{axiom}
\section{\pending{Equivalence Relation}}% \section{\defined{Equivalence Relation}}%
\label{ref:equivalence-relation} \label{ref:equivalence-relation}
Relation $R$ is an \textbf{equivalence relation} on set $A$ if and only if Relation $R$ is an \textbf{equivalence relation} on set $A$ if and only if
$R$ is a binary \nameref{ref:relation} that is \nameref{ref:reflexive} on $A$, $R$ is a binary \nameref{ref:relation} that is \nameref{ref:reflexive} on $A$,
\nameref{ref:symmetric}, and \nameref{ref:transitive}. \nameref{ref:symmetric}, and \nameref{ref:transitive}.
\begin{definition}
\lean*{Init/Core}{Equivalence}
\end{definition}
\section{\defined{Extensionality Axiom}}% \section{\defined{Extensionality Axiom}}%
\label{ref:extensionality-axiom} \label{ref:extensionality-axiom}
@ -209,7 +233,7 @@ For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
\end{axiom} \end{axiom}
\section{\pending{Partition}}% \section{\defined{Partition}}%
\label{ref:partition} \label{ref:partition}
A \textbf{partition} $\Pi$ of a set $A$ is a set of nonempty subsets of $A$ that A \textbf{partition} $\Pi$ of a set $A$ is a set of nonempty subsets of $A$ that
@ -219,6 +243,12 @@ A \textbf{partition} $\Pi$ of a set $A$ is a set of nonempty subsets of $A$ that
\item each element of $A$ is in some set in $\Pi$. \item each element of $A$ is in some set in $\Pi$.
\end{enumerate} \end{enumerate}
\begin{definition}
\lean*{Mathlib/Data/Setoid/Partition}{Setoid.IsPartition}
\end{definition}
\section{\defined{Power Set}}% \section{\defined{Power Set}}%
\label{ref:power-set} \label{ref:power-set}
@ -243,7 +273,7 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
\end{axiom} \end{axiom}
\section{\pending{Quotient Set}}% \section{\defined{Quotient Set}}%
\label{ref:quotient-set} \label{ref:quotient-set}
If $R$ is an \nameref{ref:equivalence-relation} on set $A$, then we can define If $R$ is an \nameref{ref:equivalence-relation} on set $A$, then we can define
@ -251,6 +281,12 @@ If $R$ is an \nameref{ref:equivalence-relation} on set $A$, then we can define
the equivalence classes. the equivalence classes.
The expression $A / R$ is read "$A$ modulo $R$. The expression $A / R$ is read "$A$ modulo $R$.
\begin{definition}
\lean*{Init/Core}{Quotient}
\end{definition}
\section{\defined{Range}}% \section{\defined{Range}}%
\label{ref:range} \label{ref:range}
@ -263,12 +299,18 @@ The \textbf{range} of set $R$, denoted $\ran{R}$, is given by
\end{definition} \end{definition}
\section{\pending{Reflexive}}% \section{\defined{Reflexive}}%
\label{ref:reflexive} \label{ref:reflexive}
A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for all A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for all
$x \in A$. $x \in A$.
\begin{definition}
\lean*{Init/Core}{Equivalence.refl}
\end{definition}
\section{\defined{Relation}}% \section{\defined{Relation}}%
\label{ref:relation} \label{ref:relation}
@ -305,12 +347,18 @@ For each formula $\phi$ not containing $B$, the following is an axiom:
\end{axiom} \end{axiom}
\section{\pending{Symmetric}}% \section{\defined{Symmetric}}%
\label{ref:symmetric} \label{ref:symmetric}
A binary relation $R$ is \textbf{symmetric} if and only if whenever $xRy$ then A binary relation $R$ is \textbf{symmetric} if and only if whenever $xRy$ then
$yRx$. $yRx$.
\begin{definition}
\lean*{Init/Core}{Equivalence.symm}
\end{definition}
\section{\defined{Symmetric Difference}}% \section{\defined{Symmetric Difference}}%
\label{ref:symmetric-difference} \label{ref:symmetric-difference}
@ -323,12 +371,18 @@ The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set
\end{definition} \end{definition}
\section{\pending{Transitive}}% \section{\defined{Transitive}}%
\label{ref:transitive} \label{ref:transitive}
A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and
$yRz$, then $xRz$. $yRz$, then $xRz$.
\begin{definition}
\lean*{Init/Core}{Equivalence.trans}
\end{definition}
\section{\defined{Union Axiom}}% \section{\defined{Union Axiom}}%
\label{ref:union-axiom} \label{ref:union-axiom}
@ -4406,32 +4460,82 @@ Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.28}}% \subsection{\pending{Exercise 3.28}}%
\label{sub:exercise-3.28} \label{sub:exercise-3.28}
Assume that $f$ is a one-to-one function from $A$ into $B$, and that $G$ is the 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 function with $\dom{G} = \powerset{A}$ defined by the equation
$G(X) = \img{f}{x}$. $G(X) = \img{f}{X}$.
Show that $G$ maps $\powerset{A}$ one-to-one into $\powerset{B}$. Show that $G$ maps $\powerset{A}$ one-to-one into $\powerset{B}$.
\begin{proof} \begin{proof}
TODO By construction, $\dom{G} = \powerset{A}$.
Likewise, $\ran{G} \subseteq \powerset{B}$ by definition of the
\nameref{ref:image} of sets.
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} \end{proof}
\subsection{\sorry{Exercise 3.29}}% \subsection{\pending{Exercise 3.29}}%
\label{sub:exercise-3.29} \label{sub:exercise-3.29}
Assume that $f \colon A \rightarrow B$ and define a function Assume that $f \colon A \rightarrow B$ and define a function
$G \colon B \rightarrow \powerset{A}$ by $G \colon B \rightarrow \powerset{A}$ by
$$G(b) = \{x \in A \mid f(x) = b\}.$$ \begin{equation}
\label{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. Show that if $f$ maps $A$ \textit{onto} $B$, then $G$ is one-to-one.
Does the converse hold? Does the converse hold?
\begin{proof} \begin{proof}
TODO 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 $x_2, x_2 \in \ran{f}$.
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} \end{proof}

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@ -285,7 +285,6 @@ theorem one_to_one_self_iff_one_to_one_inv {R : Relation α}
· intro ⟨hx, hy⟩ · intro ⟨hx, hy⟩
exact ⟨hy, hx⟩ exact ⟨hy, hx⟩
/-! ## Composition -/ /-! ## Composition -/
/-- /--