Enderton. Exercise 3.31, Theorem 3K, and corollary.

finite-set-exercises
Joshua Potter 2023-07-04 13:13:57 -06:00
parent 86abd77523
commit 993e9fe981
2 changed files with 296 additions and 28 deletions

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@ -44,8 +44,18 @@ For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
\end{axiom}
\section{\pending{Cartesian Product}}%
\label{ref:cartesian-product}
Let $I$ be a set and let $H$ be a \nameref{ref:function} whose domain includes $I$.
Then for each $i$ in $I$ we have the set $H(i)$.
We define the \textbf{cartesian product} of the $H(i)$'s as
$$\bigtimes_{i \in I} H(i) = \{f \mid
f \text{ is a function with domain } I \text{ and }
(\forall i \in I) f(i) \in H(i)\}.$$
\section{\defined{Compatible}}%
\label{sec:compatible}
\label{ref:compatible}
A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and
only if for all $x$ and $y$ in $A$,
@ -2569,9 +2579,8 @@ If not, then under what conditions does equality hold?
\end{proof}
\subsection{\verified{Cartesian Product}}%
\subsection{\verified{Corollary 3C}}%
\label{sub:corollary-3c}
\label{sub:cartesian-product}
\begin{theorem}[3C]
@ -3018,7 +3027,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof}
\subsection{\pending{Theorem 3K(a)}}%
\subsection{\verified{Theorem 3K(a)}}%
\label{sub:theorem-3k-a}
\begin{theorem}[3K(a)]
@ -3040,6 +3049,9 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3k\_a}
Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets.
We prove (i) \eqref{sub:theorem-3k-a-eq1} and (ii)
\eqref{sub:theorem-3k-a-eq2}.
@ -3086,7 +3098,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof}
\subsection{\pending{Theorem 3K(b)}}%
\subsection{\verified{Theorem 3K(b)}}%
\label{sub:theorem-3k-b}
\begin{theorem}[3K(b)]
@ -3103,13 +3115,23 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\img{F}{\bigcap\mathscr{A}} \subseteq
\bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}.
\end{equation}
for nonempty $\mathscr{A}$.
Equality holds if $F$ is single-rooted.
\end{theorem}
\begin{proof}
Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets.
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3k\_b\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3k\_b\_ii}
Let $F$, $A$, $B$ be arbitrary sets.
Let $\mathscr{A}$ be a nonempty set.
We first prove (i) \eqref{sub:theorem-3k-b-eq1} and (ii)
\eqref{sub:theorem-3k-b-eq2}.
Then, assuming $F$ is single-rooted, we prove both (iii)
@ -3162,14 +3184,14 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
Then $\forall A \in \mathscr{A}, v \in \img{F}{A}$.
By definition of the \nameref{ref:image} of a set,
$\forall A \in \mathscr{A}, \exists u \in A, uFv$.
Since $F$ is single-rooted, it follows that
$\exists u, \forall A \in \mathscr{A}, u \in A \land uFv$.
Since $F$ is single-rooted and $\mathscr{A}$ is nonempty, it follows that
$\exists u, (\forall A \in \mathscr{A}, u \in A) \land uFv$.
Equivalently, $\exists u \in \bigcap{A}, uFv$.
Thus $v \in \img{F}{\bigcap{A}}$.
\end{proof}
\subsection{\pending{Theorem 3K(c)}}%
\subsection{\verified{Theorem 3K(c)}}%
\label{sub:theorem-3k-c}
\begin{theorem}[3K(c)]
@ -3186,6 +3208,14 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3k\_c\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3k\_c\_ii}
We prove that (i) \eqref{sub:theorem-3k-c-eq1} holds and (ii) equality holds
if $F$ is single-rooted.
@ -3219,21 +3249,21 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof}
\subsection{\pending{Corollary 3L}}%
\subsection{\verified{Corollary 3L}}%
\label{sub:corollary-3l}
\begin{theorem}[3L]
For any function $G$ and sets $A$, $B$, and $\mathscr{A}$:
\begin{align}
G^{-1}\left\llbracket\bigcup\mathscr{A}\right\rrbracket
& = \bigcup\;\{G^{-1}[A] \mid A \in \mathscr{A}\},
\img{G^{-1}}{\bigcup{\mathscr{A}}}
& = \bigcup\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\},
\label{sub:corollary-3l-eq1} \\
G^{-1}\left[\bigcap\mathscr{A}\right]
& = \bigcap\;\{G^{-1}[A] \mid A \in \mathscr{A}\}
\img{G^{-1}}{\bigcap{\mathscr{A}}}
& = \bigcap\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\}
\text{ for } \mathscr{A} \neq \emptyset,
\label{sub:corollary-3l-eq2} \\
G^{-1}[A - B] & = G^{-1}[A] - G^{-1}[B].
\img{G^{-1}}{A - B} & = \img{G^{-1}}{A} - \img{G^{-1}}{B}.
\label{sub:corollary-3l-eq3}
\end{align}
@ -3241,6 +3271,17 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.corollary\_3l\_i}
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.corollary\_3l\_ii}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.corollary\_3l\_iii}
\nameref{sub:theorem-3k-a} implies \eqref{sub:corollary-3l-eq1}.
Because the inverse of a function is always single-rooted,
\nameref{sub:theorem-3k-b} implies \eqref{sub:corollary-3l-eq2}.
@ -3441,7 +3482,7 @@ Show that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
{Enderton.Set.Chapter\_3.exercise\_3\_2a}
Let $A$, $B$, and $C$ be arbitrary sets.
Then by definition of the \nameref{sub:cartesian-product} and union of sets,
Then by \nameref{sub:corollary-3c} and the definition of the union of sets,
\begin{align*}
A \times (B \cup C)
& = \{ \left< x, y \right> \mid x \in A \land y \in (B \cup C) \} \\
@ -3467,7 +3508,7 @@ Show that if $A \times B = A \times C$ and $A \neq \emptyset$, then $B = C$.
{Enderton.Set.Chapter\_3.exercise\_3\_2b}
Let $A$, $B$, and $C$ be arbitrary sets such that $A \neq \emptyset$.
By definition of the \nameref{sub:cartesian-product},
By \nameref{sub:corollary-3c},
\begin{align}
A \times B & = \{ \left< x, y \right> \mid x \in A \land y \in B \}
& \label{sub:exercise-3.2b-eq1} \\
@ -3513,7 +3554,7 @@ Show that $A \times \bigcup \mathscr{B} =
{Enderton.Set.Chapter\_3.exercise\_3\_3}
Let $A$ and $\mathscr{B}$ be arbitrary sets.
By definition of the \nameref{sub:cartesian-product} and the union of sets,
By \nameref{sub:corollary-3c} and the definition of the union of sets,
\begin{align*}
A \times \bigcup\mathscr{B}
& = \{ \left< x, y \right> \mid
@ -3564,10 +3605,10 @@ In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set.
Let $a \in A$.
By the \nameref{ref:pairing-axiom}, $\{a\}$ is a set.
By \nameref{sub:cartesian-product}, $\{a\} \times B$ is a set.
By \nameref{sub:corollary-3c}, $\{a\} \times B$ is a set.
Again by the \nameref{ref:pairing-axiom}, $\{\{a\} \times B\}$ is a set.
Next, by another application of \nameref{sub:cartesian-product}, $A \times B$
Next, by another application of \nameref{sub:corollary-3c}, $A \times B$
is a set.
By the \nameref{ref:power-set-axiom}, $\powerset{(A \times B)}$ is a set.
Thus, by the \nameref{ref:subset-axioms}, the following is also a set:
@ -3593,7 +3634,7 @@ In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set.
\paragraph{($\Leftarrow$)}%
Suppose $y = \{a\} \times B$ for some $a \in A$.
By \nameref{sub:cartesian-product}, $x \in \{a\} \times B$ if and only if
By \nameref{sub:corollary-3c}, $x \in \{a\} \times B$ if and only if
$\exists b \in B$ such that $x = \left< a, b \right>$.
But then $x \in y$ if and only if $\exists b \in B$ such that
$x = \left< a, b \right>$.
@ -3618,7 +3659,7 @@ With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$.
A \times B = \bigcup\; \{\{x\} \times B \mid x \in A\}.
\end{equation}
The left-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of
\nameref{sub:cartesian-product}.
\nameref{sub:corollary-3c}.
The right-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of
\nameref{sub:exercise-3.5a}.
We prove the set on each side is a subset of the other.
@ -4193,9 +4234,8 @@ Show that $F \restriction A = F \cap (A \times \ran{F})$.
\begin{proof}
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
By \nameref{sub:corollary-3c} and definition of the \nameref{ref:restriction},
intersection, and \nameref{ref:range} of sets,
\begin{align*}
F \restriction A
& = \{\left< u, v \right> \mid uFv \land u \in A\} \\
@ -4559,7 +4599,7 @@ Define
\end{proof}
\subsection{\sorry{Exercise 3.31}}%
\subsection{\unverified{Exercise 3.31}}%
\label{sub:exercise-3.31}
Show that from the first form of the axiom of choice we can prove the second
@ -4567,7 +4607,55 @@ Show that from the first form of the axiom of choice we can prove the second
\begin{proof}
TODO
We prove the first form holds if and only if the second form holds.
\paragraph{($\Rightarrow$)}%
We assume the first form of the axiom of choice.
Let $I$ be a set and $H$ be a function with $\dom{H} = I$.
Furthermore, suppose $H(i) \neq \emptyset$ for all $i \in I$.
By definition of the \nameref{ref:cartesian-product},
$$\bigtimes_{i \in I} H(i) = \{f \mid
f \text{ is a function with } \dom{f} = I \text{ and }
(\forall i \in I) f(i) \in H(i)\}.$$
Consider the relation $R$ formed by
$$R = \bigcup_{i \in I} \{i\} \times H(i).$$
By the \nameref{ref:axiom-of-choice-1}, there exists a function
$f \subseteq R$ with $\dom{f} = I$.
Furthermore, for all $i \in I$, it must be $f(i) \in H(i)$ by construction.
Then $f$ is a member of $\bigtimes_{i \in I} H(i)$.
That is, $\bigtimes_{i \in I} H(i) \neq \emptyset$.
\paragraph{($\Leftarrow$)}%
We assume the second form of the axiom of choice.
Let $R$ be an arbitrary relation.
There are two cases to consider:
\subparagraph{Case 1}%
Suppose $\ran{R} = \emptyset$.
Then $R = \emptyset$.
Thus the function $\emptyset \subseteq R$ satisfies
$\dom{\emptyset} = \dom{R}$.
\subparagraph{Case 2}%
Suppose $\ran{R} \neq \emptyset$.
Let $I = \dom{R}$ and define $H \colon I \rightarrow \{\ran{R}\}$ as
$H(i) = \ran{R}$ for all $i \in I$.
By the \nameref{ref:axiom-of-choice-2},
$\bigtimes_{i \in I} H(i) \neq \emptyset$.
By definition of the \nameref{ref:cartesian-product}, there exists some
function $f$ such that $\dom{f} = I$ and
$(\forall i \in I) f(i) \in H(i) = \ran{R}$.
Thus $\dom{f} = \dom{R}$ and $f \subseteq R$ as desired.
\paragraph{Conclusion}%
The above cases are exhaustive and yield the same conclusion: for any
relation $R$ there exists a function $f \subseteq R$ such that
$\dom{f} = \dom{R}$.
\end{proof}

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@ -534,6 +534,186 @@ theorem theorem_3j_b {F : Set.Relation α} {A B : Set α}
(∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by
sorry
/-- #### Theorem 3K (a)
The following hold for any sets. (`F` need not be a function.)
The image of a union is the union of the images:
```
F⟦ 𝓐⟧ = {F⟦A⟧ | A ∈ 𝓐}
```
-/
theorem theorem_3k_a {F : Set.Relation α} {𝓐 : Set (Set α)}
: F.image (⋃₀ 𝓐) = ⋃₀ { F.image A | A ∈ 𝓐 } := by
rw [Set.Subset.antisymm_iff]
apply And.intro
· show ∀ v, v ∈ F.image (⋃₀ 𝓐) → v ∈ ⋃₀ { F.image A | A ∈ 𝓐 }
intro v hv
unfold image at hv
simp only [Set.mem_sUnion, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
have ⟨A, hA⟩ := hu.left
simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and]
refine ⟨A, hA.left, ?_⟩
show v ∈ F.image A
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hA.right, hu.right⟩
· show ∀ v, v ∈ ⋃₀ {x | ∃ A, A ∈ 𝓐 ∧ F.image A = x} → v ∈ F.image (⋃₀ 𝓐)
intro v hv
simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and] at hv
have ⟨A, hA⟩ := hv
unfold image at hA
simp only [Set.mem_setOf_eq] at hA
have ⟨u, hu⟩ := hA.right
unfold image
simp only [Set.mem_sUnion, Set.mem_setOf_eq]
exact ⟨u, ⟨A, hA.left, hu.left⟩, hu.right⟩
/-! #### Theorem 3K (b)
The following hold for any sets. (`F` need not be a function.)
The image of an intersection is included in the intersection of the images:
```
F⟦⋂ 𝓐⟧ ⊆ ⋂ {F⟦A⟧ | A ∈ 𝓐}
```
Equality holds if `F` is single-rooted.
-/
theorem theorem_3k_b_i {F : Set.Relation α} {𝓐 : Set (Set α)}
: F.image (⋂₀ 𝓐) ⊆ ⋂₀ { F.image A | A ∈ 𝓐} := by
show ∀ v, v ∈ F.image (⋂₀ 𝓐) → v ∈ ⋂₀ { F.image A | A ∈ 𝓐}
intro v hv
unfold image at hv
simp only [Set.mem_sInter, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
simp only [
Set.mem_sInter,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
]
intro A hA
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hu.left A hA, hu.right⟩
theorem theorem_3k_b_ii {F : Set.Relation α} {𝓐 : Set (Set α)}
(hF : F.isSingleRooted) (h𝓐 : Set.Nonempty 𝓐)
: F.image (⋂₀ 𝓐) = ⋂₀ { F.image A | A ∈ 𝓐} := by
rw [Set.Subset.antisymm_iff]
refine ⟨theorem_3k_b_i, ?_⟩
show ∀ v, v ∈ ⋂₀ {x | ∃ A, A ∈ 𝓐 ∧ image F A = x} → v ∈ image F (⋂₀ 𝓐)
intro v hv
simp only [
Set.mem_sInter,
Set.mem_setOf_eq,
forall_exists_index,
and_imp,
forall_apply_eq_imp_iff₂
] at hv
unfold image at hv
simp only [Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ : ∃ u, (∀ (a : Set α), a ∈ 𝓐 → u ∈ a) ∧ (u, v) ∈ F := by
have ⟨A, hA⟩ := h𝓐
have ⟨_, ⟨_, hv'⟩⟩ := hv A hA
have ⟨u, hu⟩ := hF v (mem_pair_imp_snd_mem_ran hv')
simp only [and_imp] at hu
refine ⟨u, ?_, hu.left.right⟩
intro a ha
have ⟨u₁, hu₁⟩ := hv a ha
have := hu.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
rw [← this]
exact hu₁.left
unfold image
simp only [Set.mem_sInter, Set.mem_setOf_eq]
exact ⟨u, hu⟩
/-! #### Theorem 3K (c)
The following hold for any sets. (`F` need not be a function.)
The image of a difference includes the difference of the images:
```
F⟦A⟧ - F⟦B⟧ ⊆ F⟦A - B⟧.
```
Equality holds if `F` is single-rooted.
-/
theorem theorem_3k_c_i {F : Set.Relation α} {A B : Set α}
: F.image A \ F.image B ⊆ F.image (A \ B) := by
show ∀ v, v ∈ F.image A \ F.image B → v ∈ F.image (A \ B)
intro v hv
have hv' : v ∈ image F A ∧ v ∉ image F B := hv
conv at hv' => arg 1; unfold image; simp only [Set.mem_setOf_eq, eq_iff_iff]
have ⟨u, hu⟩ := hv'.left
have hw : ∀ w ∈ B, (w, v) ∉ F := by
intro w hw nw
have nv := hv'.right
unfold image at nv
simp only [Set.mem_setOf_eq, not_exists, not_and] at nv
exact absurd nw (nv w hw)
have hu' : u ∉ B := by
by_contra nu
exact absurd hu.right (hw u nu)
unfold image
simp only [Set.mem_diff, Set.mem_setOf_eq]
exact ⟨u, ⟨hu.left, hu'⟩, hu.right⟩
theorem theorem_3k_c_ii {F : Set.Relation α} {A B : Set α}
(hF : F.isSingleRooted)
: F.image A \ F.image B = F.image (A \ B) := by
rw [Set.Subset.antisymm_iff]
refine ⟨theorem_3k_c_i, ?_⟩
show ∀ v, v ∈ image F (A \ B) → v ∈ image F A \ image F B
intro v hv
unfold image at hv
simp only [Set.mem_diff, Set.mem_setOf_eq] at hv
have ⟨u, hu⟩ := hv
have hv₁ : v ∈ F.image A := by
unfold image
simp only [Set.mem_setOf_eq]
exact ⟨u, hu.left.left, hu.right⟩
have hv₂ : v ∉ F.image B := by
intro nv
unfold image at nv
simp only [Set.mem_setOf_eq] at nv
have ⟨u₁, hu₁⟩ := nv
have ⟨x, hx⟩ := hF v (mem_pair_imp_snd_mem_ran hu.right)
simp only [and_imp] at hx
have hr₁ := hx.right u (mem_pair_imp_fst_mem_dom hu.right) hu.right
have hr₂ := hx.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
rw [hr₂, ← hr₁] at hu₁
exact absurd hu₁.left hu.left.right
exact ⟨hv₁, hv₂⟩
/-! #### Corollary 3L
For any function `G` and sets `A`, `B`, and `𝓐`:
```
G⁻¹⟦ 𝓐⟧ = {G⁻¹⟦A⟧ | A ∈ 𝓐},
G⁻¹⟦𝓐⟧ = ⋂ {G⁻¹⟦A⟧ | A ∈ 𝓐} for 𝓐 ≠ ∅,
G⁻¹⟦A - B⟧ = G⁻¹⟦A⟧ - G⁻¹⟦B⟧.
```
-/
theorem corollary_3l_i {G : Set.Relation α} {𝓐 : Set (Set α)}
: G.inv.image (⋃₀ 𝓐) = ⋃₀ {G.inv.image A | A ∈ 𝓐} := theorem_3k_a
theorem corollary_3l_ii {G : Set.Relation α} {𝓐 : Set (Set α)}
(hG : G.isSingleValued) (h𝓐 : Set.Nonempty 𝓐)
: G.inv.image (⋂₀ 𝓐) = ⋂₀ {G.inv.image A | A ∈ 𝓐} := by
have hG' : G.inv.isSingleRooted :=
single_valued_self_iff_single_rooted_inv.mp hG
exact theorem_3k_b_ii hG' h𝓐
theorem corollary_3l_iii {G : Set.Relation α} {A B : Set α}
(hG : G.isSingleValued)
: G.inv.image (A \ B) = G.inv.image A \ G.inv.image B := by
have hG' : G.inv.isSingleRooted :=
single_valued_self_iff_single_rooted_inv.mp hG
exact (theorem_3k_c_ii hG').symm
end
end Enderton.Set.Chapter_3