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Enderton. Exercise 3.38.

finite-set-exercises
Joshua Potter 2023-07-11 16:48:46 -06:00
parent 2199503f70
commit 5991554fcb
3 changed files with 197 additions and 109 deletions
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116
Bookshelf/Enderton/Set.tex
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@ -112,7 +112,7 @@ There is a set having no members:
\end{axiom}
\section{\defined{Equivalence Class}}%
\label{sec:equivalence-class}
\label{ref:equivalence-class}
The set $[x]_R$ is defined by $$[x]_R = \{t \mid xRt\}.$$
If $R$ is an \nameref{ref:equivalence-relation} and $x \in fld R$, then $[x]_R$
@ -3409,8 +3409,7 @@ 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\_3p}
\lean{Bookshelf/Enderton/Set/Relation}{Set.Relation.modEquiv\_partition}
Let $\Pi = \{[x]_R \mid x \in A\}$.
We show that (i) there are no empty sets in $\Pi$, (ii) no two different sets
@ -5165,7 +5164,7 @@ Show that $R_\Pi$ is an equivalence relation on $A$.
\end{proof}
\subsection{\pending{Exercise 3.38}}%
\subsection{\verified{Exercise 3.38}}%
\label{sub:exercise-3.38}
\nameref{sub:theorem-3p} shows that $A / R$ is a partition of $A$ whenever $R$
@ -5175,26 +5174,90 @@ Show that if we start with the equivalence relation $R_\Pi$ of the preceding
\begin{proof}
If $\Pi$ is empty, its trivial to see $A / R_\Pi$ is likewise empty.
Thus assume $\Pi$ is not empty.
Let $C \in \Pi$.
By definition of a \nameref{ref:partition}, $C$ is nonempty.
Let $x \in C$.
By definition of a \nameref{ref:partition}, every member of $\Pi$ is disjoint
from every other.
Thus
\lean{Bookshelf/Enderton/Set/Relation}
{Enderton.Set.Chapter\_3.exercise\_3\_38}
By definition,
\begin{equation}
\label{sub:exercise-3.38-eq1}
R_\Pi = \{ (x, y) \mid (\exists B \in \Pi)(x \in B \land y \in B) \}.
\end{equation}
We prove that $A / R_\Pi = \Pi$.
By the \nameref{ref:extensionality-axiom}, these two sets are equal when
$$B \in A / R_\Pi \iff B \in \Pi.$$
We prove both directions of this biconditional.
\paragraph{($\Rightarrow$)}%
Suppose $B \in A / R_\Pi$.
By \nameref{sub:exercise-3.37}, $R_\Pi$ is an equivalence class.
Then, by definition of a \nameref{ref:quotient-set},
$$A / R_\Pi = \{[x]_{R_\Pi} \mid x \in A\},$$
whose members are the \nameref{ref:equivalence-class}es.
Thus there exists some $x \in A$ such that $B = [x]_{R_\Pi}$.
By definition of a \nameref{ref:partition}, there exists a unique set
$B' \in \Pi$ containing $x$.
Thus it suffices to prove that $B = B'$, for then $B = [x]_{R_\Pi} = B'$ is
a member of $\Pi$ as desired.
We proceed by extensionality again; that is, we show
$$y \in B \iff y \in B'.$$
\subparagraph{($\rightarrow$)}%
Suppose $y \in B$.
Then
\begin{align*}
C
& = \{t \mid (\exists B \in \Pi)(x \in B \land t \in B)\} \\
& = \{t \mid x R_\Pi t\} & \textref{sub:exercise-3.37} \\
y
& \in B = [x]_{R_\Pi} \\
& = \{t \mid \pair{x, t} \in R_\Pi\} \\
& = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\}.
& \eqref{sub:exercise-3.38-eq1}
\end{align*}
Thus there exists some $B_1 \in \Pi$ such that $x \in B_1$ and
$y \in B_1$.
By definition of a \nameref{ref:partition}, $B_1 \in \Pi$ is the unique
member of $\Pi$ containing $y$.
Thus $B_1 = B'$ meaning $y \in B'$ as desired.
\subparagraph{($\leftarrow$)}%
Suppose $y \in B'$.
By construction, $x \in B'$.
Then there exists a set $B_1$ such that $x \in B_1$ and $y \in B_1$,
namely $B'$.
Therefore
\begin{align*}
y
& \in \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\
& = \{t \mid \pair{x, t} \in R_\Pi\} \\
& = [x]_{R_\Pi} = B.
\end{align*}
\subparagraph{Conclusion}%
By the \nameref{ref:extensionality-axiom}, it follows $B = B'$.
Since $B' \in P$, it also follows $B \in P$.
\paragraph{($\Leftarrow$)}%
Let $B \in \Pi$.
By definition of a \nameref{ref:partition}, $B$ is nonempty.
Let $x \in B$.
By definition of a set, $B = \{t \mid x \in B \land t \in B\}$.
By definition of a \nameref{ref:partition}, every member of $B$ must belong
to only $B$ (i.e. no other sets in the partition).
Thus we can equivalently write
\begin{align*}
B
& = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\
& = \{ t \mid \pair{x, t} \in R_\Pi \} \\
& = [x]_{R_\Pi}.
\end{align*}
Therefore every member of $\Pi$ is of form $[x]_{R_\Pi}$ for some $x \in A$.
In other words, $$A / R_\Pi = \{[x]_{R_\Pi} \mid x \in A\} = \Pi.$$
Therefore $B \in A / R_{\Pi}$.
\end{proof}
\subsection{\pending{Exercise 3.39}}%
\subsection{\sorry{Exercise 3.39}}%
\label{sub:exercise-3.39}
Assume that we start with an equivalence relation $R$ on $A$ and define $\Pi$ to
@ -5203,22 +5266,7 @@ Show that $R_\Pi$, as defined in Exercise 37, is just $R$.
\begin{proof}
If $\Pi$ is empty, it's trivial to show $R_\Pi$ is also empty.
Thus assume $\Pi$ is not empty.
Let $C \in \Pi = A / R$.
By definition of a \nameref{ref:partition}, $C$ is nonempty.
Let $x \in C$.
Thus
\begin{align*}
C
& = [x]_R & \textref{sub:theorem-3p} \\
& = \{t \mid xRt\} \\
& = \{t \mid (\exists B \in A / R)(x \in B \land t \in B)\} \\
& = \{t \mid (\exists B \in \Pi)(x \in B \land t \in B)\} \\
& = \{t \mid x R_\Pi t\}.
\end{align*}
Thus $\pair{x, y} \in R_\Pi$ if and only if $\pair{x, y} \in R$.
By the \nameref{ref:extensionality-axiom}, $R_\Pi = R$.
TODO
\end{proof}

161
Bookshelf/Enderton/Set/Chapter_3.lean
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@ -1702,58 +1702,6 @@ theorem theorem_3m {R : Set.Relation α}
have := hS ht
exact hT this ht
/-- #### Theorem 3P
Assume that `R` is an equivalence relation on `A`. Then the set
`{[x]_R | x ∈ A}` of all equivalence classes is a partition of `P`.
-/
theorem theorem_3p {R : Set.Relation α} {A : Set α} {P : Set (Set α)}
(hR : R.isEquivalence A) (hP : P = {neighborhood R x | x ∈ A})
: isPartition P A := by
refine ⟨?_, ?_, ?_, ?_⟩
· -- Every member is a subset of `A`.
intro p hp
rw [hP] at hp
simp only [Set.mem_setOf_eq] at hp
have ⟨x, hx⟩ := hp
rw [← hx.right]
show ∀ t, t ∈ neighborhood R x → t ∈ A
intro t ht
exact hR.b_on (Or.inr (mem_pair_imp_snd_mem_ran ht))
· -- Every member is nonempty.
intro p hp
rw [hP] at hp
have ⟨x, hx⟩ := hp
rw [← hx.right]
exact ⟨x, hR.refl x hx.left⟩
· -- Every pair of members is disjoint.
intro xR hxR yR hyR h
by_contra nh
have nh' : Set.Nonempty (xR ∩ yR) := by
rw [← Set.nmem_singleton_empty]
exact nh
have ⟨z, hz⟩ := nh'
rw [hP] at hxR hyR
have ⟨x, hx⟩ := hxR
have ⟨y, hy⟩ := hyR
rw [← hx.right, ← hy.right] at hz
unfold neighborhood at hz
simp only [Set.mem_inter_iff, Set.mem_setOf_eq] at hz
have hzy : (z, y) ∈ R := hR.symm hz.right
have hxy : (x, y) ∈ R := hR.trans hz.left hzy
have := (neighborhood_iff_mem hR hx.left hy.left).mpr hxy
rw [hx.right, hy.right] at this
exact absurd this h
· -- Every element of `A` is in `P`.
intro x hx
have := hR.refl x hx
refine ⟨neighborhood R x, ?_, this⟩
· rw [hP]
exact ⟨x, hx, rfl⟩
/-- #### Exercise 3.32 (a)
Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`.
@ -1980,26 +1928,27 @@ theorem exercise_3_36 {f : Set.HRelation α β}
/-- #### Exercise 3.37
Assume that `Π` is a partition of a set `A`. Define the relation `R` as follows:
Assume that `Π` is a partition of a set `A`. Define the relation `Rₚ` as
follows:
```
xRy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
xRₚy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
```
Show that `R` is an equivalence relation on `A`. (This is a formalized version
Show that `Rₚ` is an equivalence relation on `A`. (This is a formalized version
of the discussion at the beginning of this section.)
-/
theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
(hP : isPartition P A) (R : Set.Relation α)
(hR : ∀ x y, (x, y) ∈ R ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: isEquivalence R A := by
have hR' : R = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
(hP : isPartition P A) (Rₚ : Set.Relation α)
(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: isEquivalence Rₚ A := by
have hR : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
ext p
have (x, y) := p
exact hR x y
refine ⟨?_, ?_, ?_, ?_⟩
exact hRₚ x y
· -- `fld R ⊆ A`
show ∀ x, x ∈ fld R → x ∈ A
rw [hR']
refine ⟨?_, ?_, ?_, ?_⟩
· -- `fld Rₚ ⊆ A`
show ∀ x, x ∈ fld Rₚ → x ∈ A
rw [hR]
intro x hx
unfold fld dom ran at hx
simp only [
@ -2018,25 +1967,25 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
have := hP.p_subset B hB.left
exact this hB.right.right
· -- `isReflexive R A`
· -- `isReflexive Rₚ A`
intro x hx
rw [hR']
rw [hR]
simp only [Set.mem_setOf_eq, and_self]
exact hP.exhaustive x hx
· -- `isSymmetric R`
· -- `isSymmetric Rₚ`
intro x y h
rw [hR'] at h
rw [hR] at h
simp only [Set.mem_setOf_eq] at h
have ⟨B, hB⟩ := h
rw [hR']
rw [hR]
simp only [Set.mem_setOf_eq]
conv at hB => right; rw [and_comm]
exact ⟨B, hB⟩
· -- `isTransitive R`
· -- `isTransitive Rₚ`
intro x y z hx hy
rw [hR'] at hx hy
rw [hR] at hx hy
simp only [Set.mem_setOf_eq] at hx hy
have ⟨B₁, hB₁⟩ := hx
have ⟨B₂, hB₂⟩ := hy
@ -2050,10 +1999,78 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
intro h'
rw [Set.ext_iff] at h'
exact (h' y).mp ⟨hy₁, hy₂⟩
rw [hR']
rw [hR]
simp only [Set.mem_setOf_eq]
exact ⟨B₁, hB₁.left, hB₁.right.left, by rw [hB]; exact hB₂.right.right⟩
/-- #### Exercise 3.38
Theorem 3P shows that `A / R` is a partition of `A` whenever `R` is an
equivalence relation on `A`. Show that if we start with the equivalence relation
`Rₚ` of the preceding exercise, then the partition `A / Rₚ` is just `P`.
-/
theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
(hP : isPartition P A) (Rₚ : Set.Relation α)
(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
: modEquiv (exercise_3_37 hP Rₚ hRₚ) = P := by
have hR : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
ext p
have (x, y) := p
exact hRₚ x y
ext B
apply Iff.intro
· intro ⟨x, hx⟩
have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx.left
simp only at hB'
suffices B = B' by
rw [this]
exact hB'.left.left
ext y
apply Iff.intro
· intro hy
rw [← hx.right, hR] at hy
have ⟨B₁, hB₁⟩ := hy
have := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
rw [← this]
exact hB₁.right.right
· intro hy
rw [← hx.right, hR]
exact ⟨B', hB'.left.left, hB'.left.right, hy⟩
· intro hB
have ⟨x, hx⟩ := hP.nonempty B hB
have hx' : x ∈ A := hP.p_subset B hB hx
refine ⟨x, hx', Eq.symm ?_⟩
calc B
_ = {t | x ∈ B ∧ t ∈ B} := by
ext y
apply Iff.intro
· intro hy
exact ⟨hx, hy⟩
· intro hy
exact hy.right
_ = {t | ∃ B₁ ∈ P, x ∈ B₁ ∧ t ∈ B₁} := by
ext y
apply Iff.intro
· intro hy
exact ⟨B, hB, hy⟩
· intro hy
have ⟨B₁, hB₁⟩ := hy
have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx'
simp only [Set.mem_setOf_eq] at *
have : B = B₁ := by
have lhs := hB'.right B ⟨hB, hx⟩
have rhs := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
rw [lhs, rhs]
rw [this]
exact hB₁.right
_ = {t | (x, t) ∈ Rₚ } := by
rw [hR]
simp only [Set.mem_setOf_eq]
_ = neighborhood Rₚ x := rfl
end Relation
end Enderton.Set.Chapter_3

23
Bookshelf/Enderton/Set/Relation.lean
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@ -529,6 +529,14 @@ and the members of the set. It isn't standard in anyway.
-/
def neighborhood (R : Relation α) (x : α) := { t | (x, t) ∈ R }
/--
The neighborhood with some respect to an equivalence relation `R` on set `A`
and member `x` contains `x`.
-/
theorem neighborhood_self_mem {R : Set.Relation α} {A : Set α} {x : α}
(hR : isEquivalence R A) (hx : x ∈ A)
: x ∈ neighborhood R x := hR.refl x hx
/--
Assume that `R` is an equivalence relation on `A` and that `x` and `y` belong
to `A`. Then `[x]_R = [y]_R ↔ xRy`.
@ -560,6 +568,21 @@ structure isPartition (P : Set (Set α)) (A : Set α) : Prop where
disjoint : ∀ a ∈ P, ∀ b, b ∈ P → a ≠ b → a ∩ b = ∅
exhaustive : ∀ a ∈ A, ∃ p, p ∈ P ∧ a ∈ p
/--
Membership of sets within `P` is unique.
-/
theorem partition_mem_mem_eq {P : Set (Set α)} {A : Set α}
(hP : isPartition P A) (hx : x ∈ A)
: ∃! B, B ∈ P ∧ x ∈ B := by
have ⟨B, hB⟩ := hP.exhaustive x hx
refine ⟨B, hB, ?_⟩
intro B₁ hB₁
by_contra nB
have hB_disj := hP.disjoint B hB.left B₁ hB₁.left (Ne.symm nB)
rw [Set.ext_iff] at hB_disj
have := (hB_disj x).mp ⟨hB.right, hB₁.right⟩
simp at this
/--
The partition `A / R` induced by an equivalence relation `R`.
-/