Enderton. Prove out most of the set function exercises.

finite-set-exercises
Joshua Potter 2023-06-30 08:05:12 -06:00
parent d44bcd50ea
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@ -6,11 +6,6 @@
\input{../../preamble} \input{../../preamble}
\makeleancommands{../..} \makeleancommands{../..}
\newcommand{\dom}[1]{\textop{dom}{#1}}
\newcommand{\fld}[1]{\textop{fld}{#1}}
\newcommand{\ran}[1]{\textop{ran}{#1}}
\newcommand{\img}[2]{#1\left\llbracket#2\right\rrbracket}
\begin{document} \begin{document}
\header{Elements of Set Theory}{Herbert B. Enderton} \header{Elements of Set Theory}{Herbert B. Enderton}
@ -52,8 +47,8 @@ The \textbf{composition} of sets $F$ and $G$ is
\section{\defined{Domain}}% \section{\defined{Domain}}%
\label{ref:domain} \label{ref:domain}
Given \nameref{ref:relation} $R$, the \textbf{domain} of $R$, denoted $\dom{R}$, The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by
is given by $$x \in \dom{R} \iff \exists y \left< x, y \right> \in R.$$ $$x \in \dom{R} \iff \exists y \left< x, y \right> \in R.$$
\begin{definition} \begin{definition}
@ -77,7 +72,8 @@ There is a set having no members:
\label{ref:equivalence-relation} \label{ref:equivalence-relation}
Relation $R$ is an \textbf{equivalence relation} if and only if $R$ is a binary Relation $R$ is an \textbf{equivalence relation} if and only if $R$ is a binary
relation that is reflexive, symmetric, and transitive. \nameref{ref:relation} that is \nameref{ref:reflexive},
\nameref{ref:symmetric}, and \nameref{ref:transitive}.
\section{\defined{Extensionality Axiom}}% \section{\defined{Extensionality Axiom}}%
\label{ref:extensionality-axiom} \label{ref:extensionality-axiom}
@ -233,8 +229,8 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
\section{\defined{Range}}% \section{\defined{Range}}%
\label{ref:range} \label{ref:range}
Given \nameref{ref:relation} $R$, the \textbf{range} of $R$, denoted $\ran{R}$, The \textbf{range} of set $R$, denoted $\ran{R}$, is given by
is given by $$x \in \ran{R} \iff \exists t \left< t, x \right> \in R.$$ $$x \in \ran{R} \iff \exists t \left< t, x \right> \in R.$$
\begin{definition} \begin{definition}
@ -242,8 +238,8 @@ Given \nameref{ref:relation} $R$, the \textbf{range} of $R$, denoted $\ran{R}$,
\end{definition} \end{definition}
\section{\pending{Reflexive Relation}}% \section{\pending{Reflexive}}%
\label{ref:reflexive-relation} \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
$x \in A$. $x \in A$.
@ -284,8 +280,8 @@ For each formula $\phi$ not containing $B$, the following is an axiom:
\end{axiom} \end{axiom}
\section{\pending{Symmetric Relation}}% \section{\pending{Symmetric}}%
\label{ref:symmetric-relation} \label{ref:symmetric}
A binary relation $R$ is \textbf{symmetric} on $A$ if and only if whenever A binary relation $R$ is \textbf{symmetric} on $A$ if and only if whenever
$xRy$ then $yRx$. $xRy$ then $yRx$.
@ -302,8 +298,8 @@ The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set
\end{definition} \end{definition}
\section{\pending{Transitive Relation}}% \section{\pending{Transitive}}%
\label{ref:transitive-relation} \label{ref:transitive}
A binary relation $R$ is \textbf{transitive} on $A$ if and only if whenever A binary relation $R$ is \textbf{transitive} on $A$ if and only if whenever
$xRy$ and $yRz$, then $xRz$. $xRy$ and $yRz$, then $xRz$.
@ -2885,7 +2881,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\subparagraph{($\Rightarrow$)}% \subparagraph{($\Rightarrow$)}%
Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$. Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$.
All that remains is to prove $F$ is single-valued. All that remains is to prove $F$ is single-rooted.
Let $y \in \ran{F}$. Let $y \in \ran{F}$.
By definition of the \nameref{ref:range} of a function, there exists some By definition of the \nameref{ref:range} of a function, there exists some
$x$ such that $\left< x, y \right> \in F$. $x$ such that $\left< x, y \right> \in F$.
@ -3175,7 +3171,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\section{Equivalence Relations}% \section{Equivalence Relations}%
\label{sec:equivalence-relations} \label{sec:equivalence-relations}
\subsection{\sorry{Theorem 3M}}% \subsection{\pending{Theorem 3M}}%
\label{sub:theorem-3m} \label{sub:theorem-3m}
\begin{theorem}[3M] \begin{theorem}[3M]
@ -3187,7 +3183,23 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof} \begin{proof}
TODO 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} is, by definition, a
\nameref{ref:reflexive}, symmetric, and transitive relation.
Thus 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} \end{proof}
@ -3681,7 +3693,7 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
\end{answer} \end{answer}
\subsection{\sorry{Exercise 3.11}}% \subsection{\verified{Exercise 3.11}}%
\label{sub:exercise-3.11} \label{sub:exercise-3.11}
Prove the following version (for functions) of the extensionality principle: Prove the following version (for functions) of the extensionality principle:
@ -3691,11 +3703,24 @@ Then $F = G$.
\begin{proof} \begin{proof}
TODO \lean{Init/Core}{funext}
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 $\left< x, y \right> \in F$ if and only if
$\left< x, y \right> \in G$.
But this follows immediately:
\begin{align*}
\left< x, y \right> \in F
& \iff y = F(x) \land \left< x, F(x) \right> \in F \\
& \iff y = G(x) \land \left< x, G(x) \right> \in G \\
& \iff \left< x, y \right> \in G.
\end{align*}
By the \nameref{ref:extensionality-axiom}, $F = G$.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.12}}% \subsection{\pending{Exercise 3.12}}%
\label{sub:exercise-3.12} \label{sub:exercise-3.12}
Assume that $f$ and $g$ are functions and show that Assume that $f$ and $g$ are functions and show that
@ -3704,11 +3729,31 @@ Assume that $f$ and $g$ are functions and show that
\begin{proof} \begin{proof}
TODO Let $f$ and $g$ be \nameref{ref:function}s.
\paragraph{($\Rightarrow$)}%
Suppose $f \subseteq g$.
Then for all \nameref{ref:ordered-pair}s $\left< x, y \right>$,
$\left< x, y \right> \in f$ implies $\left< x, y \right> \in g$.
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 $\left< x, y \right> \in f$.
By hypothesis, $x \in \dom{g}$ and $y = f(x) = g(x)$.
Thus $\left< x, y \right> \in g$ as well.
Therefore $f \subseteq g$.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.13}}% \subsection{\pending{Exercise 3.13}}%
\label{sub:exercise-3.13} \label{sub:exercise-3.13}
Assume that $f$ and $g$ are functions with $f \subseteq g$ and Assume that $f$ and $g$ are functions with $f \subseteq g$ and
@ -3717,11 +3762,17 @@ Show that $f = g$.
\begin{proof} \begin{proof}
TODO 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} \end{proof}
\subsection{\sorry{Exercise 3.14}}% \subsection{\pending{Exercise 3.14}}%
\label{sub:exercise-3.14} \label{sub:exercise-3.14}
Assume that $f$ and $g$ are functions. Assume that $f$ and $g$ are functions.
@ -3734,35 +3785,116 @@ Assume that $f$ and $g$ are functions.
\begin{proof} \begin{proof}
TODO Assume $f$ and $g$ are functions.
\paragraph{(a)}%
Consider $f \cap g$.
By definition of the intersection of sets, $f \cap g \subseteq f$.
By \nameref{sub:exercise-3.12}, $\dom{(f \cap g)} = \dom{f}$ and
$(\forall x \in \dom{(f \cap g)} (f \cap g)(x) = f(x)$.
The latter conjunct shows that, since $f$ is single-valued, $f \cap g$ must
also be single-valued.
In other words, $f \cap g$ 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 $\left< x, y_1 \right> \in f$.
Likewise there exists only one $y_2$ such that
$\left< x, y_2 \right> \in g$.
But $\left< x, y_1 \right> \in f \cup g$ and
$\left< x, y_2 \right> \in f \cup g$.
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 $\left< x, y \right> \in f \cup g$.
\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 $\left< x, y \right> \in f \cup g$.
\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 $\left< x, y \right> \in f \cup g$.
\end{enumerate}
The above cases are exhaustive.
Together they imply that $f \cup g$ is single-valued, i.e. a function.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.15}}% \subsection{\pending{Exercise 3.15}}%
\label{sub:exercise-3.15} \label{sub:exercise-3.15}
Let $\mathscr{A}$ be a set of functions such that for any $f$ and $g$ in Let $\mathscr{A}$ be a set of functions such that for any $f$ and $g$ in
$\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$. $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$.
Show that $\bigcup \mathscr{A}$ is a function. Show that $\bigcup{\mathscr{A}}$ is a function.
\begin{proof} \begin{proof}
TODO 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
$\left< x, y_1 \right> \in \bigcup{\mathscr{A}}$.
Suppose there also exists some $y_2$ such that
$\left< x, y_2 \right> \in \bigcup{\mathscr{A}}$.
By definition of the union of sets, there exists some function
$f \in \mathscr{A}$ such that $\left< x, y_1 \right> \in f$.
Likewise there exists some function $g \in \mathscr{A}$ such that
$\left< x, y_2 \right> \in g$.
There are two cases to consider:
\paragraph{Case 1}%
Suppose $f \subseteq g$.
Then $\left< x, y_1 \right>, \left< x, y_2 \right> \in g$.
Since $g$ is a function, i.e. single-valued, $y_1 = y_2$.
\paragraph{Case 2}%
Suppose $g \subseteq f$.
Then $\left< x, y_1 \right>, \left< x, y_2 \right> \in f$.
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} \end{proof}
\subsection{\sorry{Exercise 3.16}}% \subsection{\unverified{Exercise 3.16}}%
\label{sub:exercise-3.16} \label{sub:exercise-3.16}
Show that there is no set to which every function belongs. Show that there is no set to which every function belongs.
\begin{proof} \begin{proof}
TODO 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} \end{proof}
\subsection{\sorry{Exercise 3.17}}% \subsection{\pending{Exercise 3.17}}%
\label{sub:exercise-3.17} \label{sub:exercise-3.17}
Show that the composition of two single-rooted sets is again single-rooted. Show that the composition of two single-rooted sets is again single-rooted.
@ -3770,11 +3902,36 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
\begin{proof} \begin{proof}
TODO Let $F$ and $G$ be two single-rooted sets.
Consider $F \circ G$.
By definition of the \nameref{ref:composition} of sets,
\begin{equation}
\label{sub:exercise-3.17-eq1}
F \circ G = \{\left< u, v \right> \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 $\left< u_1, v \right> \in F \circ G$.
Let $u_2$ be a set such that $\left< u_2, v \right> \in F \circ G$.
By \eqref{sub:exercise-3.17-eq1}, there exists a set $t_1$ such that
$\left< u_1, t_1 \right> \in G$ and $\left< t_1, v \right> \in F$.
Likewise, there exists a set $t_2$ such that
$\left< u_2, t_2 \right> \in G$ and $\left< t_2, v \right> \in F$.
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} \end{proof}
\subsection{\sorry{Exercise 3.18}}% \subsection{\pending{Exercise 3.18}}%
\label{sub:exercise-3.18} \label{sub:exercise-3.18}
Let $R$ be the set Let $R$ be the set
@ -3785,11 +3942,25 @@ Evaluate the following: $R \circ R$, $R \restriction \{1\}$,
\begin{proof} \begin{proof}
TODO \begin{enumerate}[(i)]
\item $R \circ R = \{
\left< 0, 2 \right>,
\left< 0, 3 \right>,
\left< 0, 3 \right>,
\left< 1, 3 \right>
\}$.
\item $R \restriction \{1\} = \{
\left< 1, 2 \right>,
\left< 1, 3 \right>
\}$.
\item $R^{-1} \restriction \{1\} = \{\left< 1, 0 \right>\}$.
\item $\img{R}{\{1\}} = \{2, 3\}$.
\item $\img{R^{-1}}{\{1\}} = \{0\}$.
\end{enumerate}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.19}}% \subsection{\pending{Exercise 3.19}}%
\label{sub:exercise-3.19} \label{sub:exercise-3.19}
Let $$A = \{ Let $$A = \{
@ -3804,33 +3975,80 @@ Evaluate each of the following: $A(\emptyset)$, $\img{A}{\emptyset}$,
\begin{proof} \begin{proof}
TODO \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 $A^{-1} = \{
\left< \{\emptyset, \{\emptyset\}\}, \emptyset \right>,
\left< \emptyset, \{\emptyset\} \right>
\}$.
\item $A \circ A =
\{\left< \{\emptyset\}, \{\emptyset, \{\emptyset\}\} \right>\}$.
\item $A \restriction \emptyset = \emptyset$
\item $A \restriction \{\emptyset\} =
\{\left< \emptyset, \{\emptyset, \{\emptyset\}\} \right>\}$.
\item $A \restriction \{\emptyset, \{\emptyset\}\} = A$.
\item $\bigcup\bigcup A =
\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
\end{enumerate}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.20}}% \subsection{\pending{Exercise 3.20}}%
\label{sub:exercise-3.20} \label{sub:exercise-3.20}
Show that $F \restriction A = F \cap (A \times \ran{F})$. Show that $F \restriction A = F \cap (A \times \ran{F})$.
\begin{proof} \begin{proof}
TODO Let $F$ and $A$ be arbitrary sets.
By definition of the \nameref{ref:restriction}, intersection,
\nameref{ref:range}, and \nameref{sub:cartesian-product} of sets,
Then
\begin{align*}
F \restriction A
& = \{\left< u, v \right> \mid uFv \land u \in A\} \\
& = \{\left< u, v \right> \mid
uFv \land u \in A \land v \in \ran{F}\} \\
& = \{\left< u, v \right> \mid uFv\} \cap
\{\left< u, v \right> \mid u \in A \land v \in \ran{F}\} \\
& = F \cap \{\left< u, v \right> \mid u \in A \land v \in \ran{F}\} \\
& = F \cap (A \times \ran{F}).
\end{align*}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.21}}% \subsection{\pending{Exercise 3.21}}%
\label{sub:exercise-3.21} \label{sub:exercise-3.21}
Show that $(R \circ S) \circ T = R \circ (S \circ T)$. Show that $(R \circ S) \circ T = R \circ (S \circ T)$.
\begin{proof} \begin{proof}
TODO Let $R$, $S$, and $T$ be arbitrary sets.
By definition of the \nameref{ref:composition} of sets,
\begin{align*}
(R \circ S) \circ T
& = \{\left< u, v \right> \mid
\exists t(u(R \circ S)t \land tTv)\} \\
& = \{\left< u, v \right> \mid
\exists t((\exists a, uRa \land aSt) \land tTv)\} \\
& = \{\left< u, v \right> \mid
\exists t, \exists a, (uRa \land aSt \land tTv)\} \\
& = \{\left< u, v \right> \mid
\exists a, \exists t, (uRa \land aSt \land tTv)\} \\
& = \{\left< u, v \right> \mid
\exists a(uRa \land (\exists t, aSt \land tTv))\} \\
& = \{\left< u, v \right> \mid \exists a(uRa \land a(S \circ T)v)\} \\
& = R \circ (S \circ T).
\end{align*}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.22}}% \subsection{\pending{Exercise 3.22}}%
\label{sub:exercise-3.22} \label{sub:exercise-3.22}
Show that the following are correct for any sets. Show that the following are correct for any sets.
@ -3844,11 +4062,55 @@ Show that the following are correct for any sets.
\begin{proof} \begin{proof}
TODO 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) \left< u, v \right> \in F \circ G\} \\
& = \{v \mid (\exists u \in A)
\left< u, v \right> \in \{\left< b, c \right> \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)
& = \{\left< u, v \right> \mid uQv \land u \in A \cup B\} \\
& = \{\left< u, v \right> \mid uQv \land (u \in A \lor u \in B)\} \\
& = \{\left< u, v \right> \mid
(uQv \land u \in A) \lor (uQv \land u \in B)\} \\
& = \{\left< u, v \right> \mid uQv \land u \in A\} \cup
\{\left< u, v \right> \mid uQv \land u \in B\} \\
& = (Q \restriction A) \cup (Q \restriction B).
\end{align*}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.23}}% \subsection{\pending{Exercise 3.23}}%
\label{sub:exercise-3.23} \label{sub:exercise-3.23}
Let $I_A$ be the identity function on the set $A$. Let $I_A$ be the identity function on the set $A$.
@ -3858,11 +4120,53 @@ Show that for any sets $B$ and $C$,
\begin{proof} \begin{proof}
TODO Let $I_A$ be the identity function on the set $A$.
That is, $I_A = \{\left< u, u \right> \mid u \in A\}$.
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 $\left< x, y \right> \in B \circ I_A$.
By definition of the \nameref{ref:composition} of sets,
there exists some $t$ such that $xBt$ and $t(I_A)y$.
By definition of the identity function, $I_A(t) = y$ implies $t = y$.
Thus $xBy$.
By hypothesis, $x \in \dom{(B \circ I_A)}$.
Therefore $x \in \dom{I_A} = A$.
Thus
$$\left< x, y \right>
\in \{\left< u, v \right> \mid u \in A \land uBv\}
= B \restriction A.$$
\subparagraph{($\supseteq$)}%
Let $\left< x, y \right> \in B \restriction A$.
By definition of the \nameref{ref:restriction} of sets,
$x \in A$ and $xBy$.
But $I_A(x) = x$ meaning $\left< I_A(x), y \right> \in B$.
In other words, $\left< x, y \right> \in B \circ I_A$.
\paragraph{(ii)}%
By definition of the \nameref{ref:image} of sets,
\begin{align*}
\img{I_A}{C}
& = \{v \mid (\exists u \in C) \left< u, v \right> \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} \end{proof}
\subsection{\sorry{Exercise 3.24}}% \subsection{\pending{Exercise 3.24}}%
\label{sub:exercise-3.24} \label{sub:exercise-3.24}
Show that for a function $F$, Show that for a function $F$,
@ -3870,11 +4174,20 @@ Show that for a function $F$,
\begin{proof} \begin{proof}
TODO 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) \left< x, y \right> \in F\} \\
& = \{x \mid \exists x \in F, F(x) \in A\} \\
& = \{x \in \dom{F} \mid F(x) \in A\}.
\end{align*}
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.25}}% \subsection{\pending{Exercise 3.25}}%
\label{sub:exercise-3.25} \label{sub:exercise-3.25}
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
@ -3887,22 +4200,38 @@ Show that for a function $F$,
\begin{proof} \begin{proof}
TODO \paragraph{(b)}%
\label{par:exercise-3.25-b}
Let $G$ be a function.
Let $\left< x, y \right> \in G \circ G^{-1}$.
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,
$$x(G^{-1})t \iff tGx.$$
By hypothesis, $G$ is single-valued.
Thus $x = y$.
That is, $G \circ G^{-1}$ is the identity function on $\ran{G}$.
\paragraph{(a)}%
This immediately follows from part \nameref{par:exercise-3.25-b}.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.26}}% \subsection{\pending{Exercise 3.26}}%
\label{sub:exercise-3.26} \label{sub:exercise-3.26}
Prove the second halves of parts (a) and (b) of Theorem 3K. Prove the second halves of parts (a) and (b) of Theorem 3K.
\begin{proof} \begin{proof}
TODO Refer to \nameref{sub:theorem-3k-a}, \nameref{sub:theorem-3k-b}, and
\nameref{sub:theorem-3k-c}.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 3.27}}% \subsection{\pending{Exercise 3.27}}%
\label{sub:exercise-3.27} \label{sub:exercise-3.27}
Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$. Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$.
@ -3910,7 +4239,36 @@ Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$.
\begin{proof} \begin{proof}
TODO 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 $\left< x, y \right> \in F \circ G$.
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,
$\left< x, t \right> \in F \circ G$.
Therefore $x \in \dom{(F \circ G)}$.
\end{proof} \end{proof}

View File

@ -507,10 +507,18 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
**iff** `F` is one-to-one. **iff** `F` is one-to-one.
-/ -/
theorem theorem_3j_a {F : Set.Relation α} {A B : Set α} theorem theorem_3j_a {F : Set.Relation α} {A B : Set α}
(hF : F.mapsInto A B) (hA : Set.Nonempty A) (hF : F.isSingleValued ∧ F.mapsInto A B) (hA : Set.Nonempty A)
: (∃ G : Set.Relation α, : (∃ G : Set.Relation α,
G.mapsInto B A ∧ (∀ p ∈ G.comp F, p.1 = p.2)) ↔ F.isOneToOne := by G.isSingleValued ∧ G.mapsInto B A ∧
(∀ p ∈ G.comp F, p.1 = p.2)) ↔ F.isOneToOne := by
apply Iff.intro
· intro ⟨G, ⟨hG₁, hG₂, hI⟩⟩
refine ⟨hF.left, ?_⟩
show F.isSingleRooted
intro y hy
have ⟨x, hx⟩ := ran_exists hy
sorry sorry
· sorry
/-- #### Theorem 3J (b) /-- #### Theorem 3J (b)
@ -519,9 +527,10 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
`B` **iff** `F` maps `A` onto `B`. `B` **iff** `F` maps `A` onto `B`.
-/ -/
theorem theorem_3j_b {F : Set.Relation α} {A B : Set α} theorem theorem_3j_b {F : Set.Relation α} {A B : Set α}
(hF : F.mapsInto A B) (hA : Set.Nonempty A) (hF : F.isSingleValued ∧ F.mapsInto A B) (hA : Set.Nonempty A)
: (∃ H : Set.Relation α, : (∃ H : Set.Relation α,
H.mapsInto B A ∧ (∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by H.isSingleValued ∧ H.mapsInto B A ∧
(∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by
sorry sorry
end end

View File

@ -131,12 +131,16 @@
\newcommand{\abs}[1]{\left|#1\right|} \newcommand{\abs}[1]{\left|#1\right|}
\newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil}
\newcommand{\dom}[1]{\textop{dom}{#1}}
\newcommand{\fld}[1]{\textop{fld}{#1}}
\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}
\newcommand{\icc}[2]{\left[#1, #2\right]} \newcommand{\icc}[2]{\left[#1, #2\right]}
\newcommand{\ico}[2]{\left[#1, #2\right)} \newcommand{\ico}[2]{\left[#1, #2\right)}
\newcommand{\img}[2]{#1\!\left\llbracket#2\right\rrbracket}
\newcommand{\ioc}[2]{\left(#1, #2\right]} \newcommand{\ioc}[2]{\left(#1, #2\right]}
\newcommand{\ioo}[2]{\left(#1, #2\right)} \newcommand{\ioo}[2]{\left(#1, #2\right)}
\newcommand{\powerset}[1]{\mathscr{P}#1} \newcommand{\powerset}[1]{\mathscr{P}#1}
\newcommand{\ran}[1]{\textop{ran}{#1}}
\newcommand{\textop}[1]{\mathop{\text{#1}}} \newcommand{\textop}[1]{\mathop{\text{#1}}}
\newcommand{\ubar}[1]{\text{\b{$#1$}}} \newcommand{\ubar}[1]{\text{\b{$#1$}}}