Enderton. Finish exercises 6.

finite-set-exercises
Joshua Potter 2023-06-22 14:34:50 -06:00
parent 8b5736397c
commit 3d556bc613
2 changed files with 202 additions and 8 deletions

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@ -2792,7 +2792,7 @@ Show that if $R$ is a relation, then $\fld{R} = \bigcup\bigcup R$.
\end{proof} \end{proof}
\subsection{\unverified{Exercise 6.8}}% \subsection{\verified{Exercise 6.8}}%
\label{sub:exercise-6.8} \label{sub:exercise-6.8}
Show that for any set $\mathscr{A}$: Show that for any set $\mathscr{A}$:
@ -2807,21 +2807,112 @@ Show that for any set $\mathscr{A}$:
\begin{proof} \begin{proof}
TODO \statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_6\_8\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_6\_8\_ii}
We prove (i) \eqref{sub:exercise-6.8-eq1} and then (ii)
\eqref{sub:exercise-6.8-eq2}.
\paragraph{(i)}%
Let $x \in \dom{\bigcup{\mathscr{A}}}$.
By definition of a domain, there exists some $y$ such that
$\left< x, y \right> \in \bigcup{\mathscr{A}}$.
By definition of the union of sets,
$\exists y, \exists R \in \mathscr{A}, \left< x, y \right> \in R$.
Equivalently,
$\exists R \in \mathscr{A}, \exists y, \left< x, y \right> \in R$.
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-6.8-eq1} holds.
\paragraph{(ii)}%
Let $x \in \ran{\bigcup{\mathscr{A}}}$.
By definition of a range, there exists some $t$ such that
$\left< t, x \right> \in \bigcup{\mathscr{A}}$.
By definition of the union of sets,
$\exists t, \exists R \in \mathscr{A}, \left< t, x \right> \in R$.
Equivalently,
$\exists R \in \mathscr{A}, \exists t, \left< t, x \right> \in R$.
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-6.8-eq2} holds.
\end{proof} \end{proof}
\subsection{\unverified{Exercise 6.9}}% \subsection{\verified{Exercise 6.9}}%
\label{sub:exercise-6.9} \label{sub:exercise-6.9}
Discuss the result of replacing the union operation by the intersection Discuss the result of replacing the union operation by the intersection
operation in the preceding problem. operation in the preceding problem.
\begin{proof} \begin{answer}
TODO \statementpadding
\end{proof} \lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_6\_9\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_6\_9\_ii}
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} \},
& \label{sub:exercise-6.9-eq1} \\
\ran{\bigcap{\mathscr{A}}}
& \subseteq \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}.
& \label{sub:exercise-6.9-eq2}
\end{align}
We prove (i) \eqref{sub:exercise-6.9-eq1} and then (ii)
\eqref{sub:exercise-6.9-eq2}.
\paragraph{(i)}%
Let $x \in \dom{\bigcap{\mathscr{A}}}$.
By definition of the domain of a set,
$\exists y, \left< x, y \right> \in \bigcap{\mathscr{A}}$.
By definition of the intersection of sets,
$\exists y, \forall R \in \mathscr{A}, \left< x, y \right> \in R$.
But this implies that
$\forall R \in \mathscr{A}, \exists y, \left< x, y \right> \in R$.
By another application of the definition of the 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-6.9-eq1} holds.
\paragraph{(ii)}%
Let $x \in \ran{\bigcap{\mathscr{A}}}$.
By definition of the range of a set,
$\exists t, \left< t, x \right> \in \bigcap{\mathscr{A}}$.
By definition of the intersection of sets,
$\exists t, \forall R \in \mathscr{A}, \left< t, x \right> \in R$.
But this implies that
$\forall R \in \mathscr{A}, \exists t, \left< t, x \right> \in R$.
By another application of the definition of the domain 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-6.9-eq2} holds.
\end{answer}
\section{Exercises 7}% \section{Exercises 7}%
\label{sec:exercises-7} \label{sec:exercises-7}

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@ -329,4 +329,107 @@ theorem exercise_6_7 {R : Set.Relation α}
simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
exact hxy_mem this exact hxy_mem this
/-- ### Exercise 6.8i
Show that for any set `𝓐`:
```
dom A = { dom R | R ∈ 𝓐 }
```
-/
theorem exercise_6_8_i {A : Set (Set.Relation α)}
: Set.Relation.dom (⋃₀ A) = ⋃₀ { Set.Relation.dom R | R ∈ A } := by
ext x
unfold Set.Relation.dom Prod.fst
simp only [
Set.mem_image,
Set.mem_sUnion,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
exists_exists_and_eq_and
]
apply Iff.intro
· intro ⟨y, ⟨t, ⟨ht, hx⟩⟩⟩
exact ⟨t, ⟨ht, ⟨y, hx⟩⟩⟩
· intro ⟨t, ⟨ht, ⟨y, hx⟩⟩⟩
exact ⟨y, ⟨t, ⟨ht, hx⟩⟩⟩
/-- ### Exercise 6.8ii
Show that for any set `𝓐`:
```
ran A = { ran R | R ∈ 𝓐 }
```
-/
theorem exercise_6_8_ii {A : Set (Set.Relation α)}
: Set.Relation.ran (⋃₀ A) = ⋃₀ { Set.Relation.ran R | R ∈ A } := by
ext x
unfold Set.Relation.ran Prod.snd
simp only [
Set.mem_image,
Set.mem_sUnion,
Prod.exists,
exists_eq_right,
Set.mem_setOf_eq,
exists_exists_and_eq_and
]
apply Iff.intro
· intro ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
exact ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
/-- ## Exercise 6.9i
Discuss the result of replacing the union operation by the intersection
operation in the preceding problem.
```
dom A = { dom R | R ∈ 𝓐 }
```
-/
theorem exercise_6_9_i {A : Set (Set.Relation α)}
: Set.Relation.dom (⋂₀ A) ⊆ ⋂₀ { Set.Relation.dom R | R ∈ A } := by
show ∀ x, x ∈ Set.Relation.dom (⋂₀ A) → x ∈ ⋂₀ { Set.Relation.dom R | R ∈ A }
unfold Set.Relation.dom Prod.fst
simp only [
Set.mem_image,
Set.mem_sInter,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro _ y hy R hR
exact ⟨y, hy R hR⟩
/-- ## Exercise 6.9ii
Discuss the result of replacing the union operation by the intersection
operation in the preceding problem.
```
ran A = { ran R | R ∈ 𝓐 }
```
-/
theorem exercise_6_9_ii {A : Set (Set.Relation α)}
: Set.Relation.ran (⋂₀ A) ⊆ ⋂₀ { Set.Relation.ran R | R ∈ A } := by
show ∀ x, x ∈ Set.Relation.ran (⋂₀ A) → x ∈ ⋂₀ { Set.Relation.ran R | R ∈ A }
unfold Set.Relation.ran Prod.snd
simp only [
Set.mem_image,
Set.mem_sInter,
Prod.exists,
exists_and_right,
exists_eq_right,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro _ y hy R hR
exact ⟨y, hy R hR⟩
end Enderton.Set.Chapter_3 end Enderton.Set.Chapter_3