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