Add prompts for upcoming theorems/exercises.

finite-set-exercises
Joshua Potter 2023-06-16 08:16:33 -06:00
parent 2abd4b6b25
commit 738435e299
2 changed files with 266 additions and 13 deletions

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

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@ -328,4 +328,22 @@ theorem theorem_3d {A : Set (Set (Set (αα)))} (h : OrderedPair x y ∈ A)
have : ∀ t, t ∈ {Sum.inl x, Sum.inr y} → t ∈ ⋃₀ (⋃₀ A) := hq'
exact ⟨this (Sum.inl x) (by simp), this (Sum.inr y) (by simp)⟩
/-- ### Exercise 6.6
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
-/
theorem exercise_6_6 {A : Set (α × β)}
: A ⊆ Set.prod (Prod.fst '' A) (Prod.snd '' A) := by
show ∀ t, t ∈ A → t ∈ Set.prod (Prod.fst '' A) (Prod.snd '' A)
intro (a, b) ht
unfold Set.prod
simp only [
Set.mem_image,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq
]
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
end Enderton.Set.Chapter_3