bookshelf/Bookshelf/Enderton/Set/Chapter_3.lean

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import Mathlib.Data.Set.Basic
import Bookshelf.Enderton.Set.Chapter_2
import Common.Logic.Basic
import Common.Set.Basic
/-! # Enderton.Chapter_3
Relations and Functions
-/
namespace Enderton.Set.Chapter_3
/-! ## Ordered Pairs -/
/--
Kazimierz Kuratowski's definition of an ordered pair.
-/
def OrderedPair (x : α) (y : β) : Set (Set (α ⊕ β)) :=
{{Sum.inl x}, {Sum.inl x, Sum.inr y}}
namespace OrderedPair
/--
For any sets `x`, `y`, `u`, and `v`, `⟨u, v⟩ = ⟨x, y⟩` **iff** `u = x ∧ v = y`.
-/
theorem ext_iff {x u : α} {y v : β}
: (OrderedPair x y = OrderedPair u v) ↔ (x = u ∧ y = v) := by
unfold OrderedPair
apply Iff.intro
· intro h
have hu := Set.ext_iff.mp h {Sum.inl u}
have huv := Set.ext_iff.mp h {Sum.inl u, Sum.inr v}
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
true_or,
iff_true
] at hu
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
or_true,
iff_true
] at huv
apply Or.elim hu
· apply Or.elim huv
· -- #### Case 1
-- `{u} = {x}` and `{u, v} = {x}`.
intro huv_x hu_x
rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_x
have hx_v := Set.pair_eq_singleton_mem_imp_eq_self huv_x
rw [hu_x, hx_v] at h
simp only [Set.mem_singleton_iff, Set.insert_eq_of_mem] at h
have := Set.pair_eq_singleton_mem_imp_eq_self $
Set.pair_eq_singleton_mem_imp_eq_self h
rw [← hx_v] at this
injection hu_x with p
injection this with q
exact ⟨p.symm, q⟩
· -- #### Case 2
-- `{u} = {x}` and `{u, v} = {x, y}`.
intro huv_xy hu_x
rw [Set.singleton_eq_singleton_iff] at hu_x
rw [hu_x] at huv_xy
by_cases hx_v : Sum.inl x = Sum.inr v
· rw [hx_v] at huv_xy
simp at huv_xy
have := Set.pair_eq_singleton_mem_imp_eq_self huv_xy.symm
injection hu_x with p
injection this with q
exact ⟨p.symm, q⟩
· rw [Set.ext_iff] at huv_xy
have := huv_xy (Sum.inr v)
simp at this
injection hu_x with p
exact ⟨p.symm, this.symm⟩
· apply Or.elim huv
· -- #### Case 3
-- `{u} = {x, y}` and `{u, v} = {x}`.
intro huv_x _
rw [Set.ext_iff] at huv_x
have hv_x := huv_x (Sum.inr v)
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
or_true,
true_iff
] at hv_x
· -- #### Case 4
-- `{u} = {x, y}` and `{u, v} = {x, y}`.
intro _ hu_xy
rw [Set.ext_iff] at hu_xy
have hy_u := hu_xy (Sum.inr y)
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
or_true,
iff_true
] at hy_u
· intro h
rw [h.left, h.right]
end OrderedPair
/-- ### Theorem 3B
If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
-/
theorem theorem_3b {C : Set (αα)} (hx : Sum.inl x ∈ C) (hy : Sum.inr y ∈ C)
: OrderedPair x y ∈ 𝒫 𝒫 C := by
have hxs : {Sum.inl x} ⊆ C := Set.singleton_subset_iff.mpr hx
have hxys : {Sum.inl x, Sum.inr y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
exact Set.mem_mem_imp_pair_subset hxs hxys
/-- ### Exercise 5.1
Suppose that we attempted to generalize the Kuratowski definitions of ordered
pairs to ordered triples by defining
```
⟨x, y, z⟩* = {{x}, {x, y}, {x, y, z}}.open Set
```
Show that this definition is unsuccessful by giving examples of objects `u`,
`v`, `w`, `x`, `y`, `z` with `⟨x, y, z⟩* = ⟨u, v, w⟩*` but with either `y ≠ v`
or `z ≠ w` (or both).
-/
theorem exercise_5_1 {x y z u v w : }
(hx : x = 1) (hy : y = 1) (hz : z = 2)
(hu : u = 1) (hv : v = 2) (hw : w = 2)
: ({{x}, {x, y}, {x, y, z}} : Set (Set )) = {{u}, {u, v}, {u, v, w}}
∧ y ≠ v := by
apply And.intro
· rw [hx, hy, hz, hu, hv, hw]
simp
· rw [hy, hv]
simp only
/-- ### Exercise 5.2a
Show that `A × (B C) = (A × B) (A × C)`.
-/
theorem exercise_5_2a {A : Set α} {B C : Set β}
: Set.prod A (B C) = (Set.prod A B) (Set.prod A C) := by
calc Set.prod A (B C)
_ = { p | p.1 ∈ A ∧ p.2 ∈ B C } := rfl
_ = { p | p.1 ∈ A ∧ (p.2 ∈ B p.2 ∈ C) } := rfl
_ = { p | (p.1 ∈ A ∧ p.2 ∈ B) (p.1 ∈ A ∧ p.2 ∈ C) } := by
ext x
rw [Set.mem_setOf_eq]
conv => lhs; rw [and_or_left]
_ = { p | p ∈ Set.prod A B (p ∈ Set.prod A C) } := rfl
_ = (Set.prod A B) (Set.prod A C) := rfl
/-- ### Exercise 5.2b
Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
-/
theorem exercise_5_2b {A : Set α} {B C : Set β}
(h : Set.prod A B = Set.prod A C) (hA : Set.Nonempty A)
: B = C := by
by_cases hB : Set.Nonempty B
· suffices B ⊆ C ∧ C ⊆ B from Set.Subset.antisymm_iff.mpr this
have ⟨a, ha⟩ := hA
apply And.intro
· show ∀ t, t ∈ B → t ∈ C
intro t ht
have : (a, t) ∈ Set.prod A B := ⟨ha, ht⟩
rw [h] at this
exact this.right
· show ∀ t, t ∈ C → t ∈ B
intro t ht
have : (a, t) ∈ Set.prod A C := ⟨ha, ht⟩
rw [← h] at this
exact this.right
· have nB : B = ∅ := Set.not_nonempty_iff_eq_empty.mp hB
rw [nB, Set.prod_right_emptyset_eq_emptyset, Set.ext_iff] at h
rw [nB]
by_contra nC
have ⟨a, ha⟩ := hA
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
exact (h (a, c)).mpr ⟨ha, hc⟩
/-- ### Exercise 5.3
Show that `A × 𝓑 = {A × X | X ∈ 𝓑}`.
-/
theorem exercise_5_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
: Set.prod A (⋃₀ 𝓑) = ⋃₀ {Set.prod A X | X ∈ 𝓑} := by
calc Set.prod A (⋃₀ 𝓑)
_ = { p | p.1 ∈ A ∧ p.2 ∈ ⋃₀ 𝓑} := rfl
_ = { p | p.1 ∈ A ∧ ∃ b ∈ 𝓑, p.2 ∈ b } := rfl
_ = { p | ∃ b ∈ 𝓑, p.1 ∈ A ∧ p.2 ∈ b } := by
ext x
rw [Set.mem_setOf_eq]
apply Iff.intro
· intro ⟨h₁, ⟨b, h₂⟩⟩
exact ⟨b, ⟨h₂.left, ⟨h₁, h₂.right⟩⟩⟩
· intro ⟨b, ⟨h₁, ⟨h₂, h₃⟩⟩⟩
exact ⟨h₂, ⟨b, ⟨h₁, h₃⟩⟩⟩
_ = ⋃₀ { Set.prod A p | p ∈ 𝓑 } := by
ext x
rw [Set.mem_setOf_eq]
unfold Set.sUnion sSup Set.instSupSetSet
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
apply Iff.intro
· intro ⟨b, ⟨h₁, ⟨h₂, h₃⟩⟩⟩
exact ⟨b, ⟨h₁, ⟨h₂, h₃⟩⟩⟩
· intro ⟨b, ⟨h₁, ⟨h₂, h₃⟩⟩⟩
exact ⟨b, ⟨h₁, ⟨h₂, h₃⟩⟩⟩
/-- ### Exercise 5.5a
Assume that `A` and `B` are given sets, and show that there exists a set `C`
such that for any `y`,
```
y ∈ C ↔ y = {x} × B for some x in A.
```
In other words, show that `{{x} × B | x ∈ A}` is a set.
-/
theorem exercise_5_5a {A : Set α} {B : Set β}
: ∃ C : Set (Set (α × β)),
y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
refine ⟨C, ?_⟩
apply Iff.intro
· intro hC
simp only [Set.mem_setOf_eq] at hC
have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
refine ⟨a, ⟨ha, ?_⟩⟩
ext x
apply Iff.intro
· intro hxy
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
rw [Prod.ext_iff] at hx
simp only at hx
rw [← hx.right] at hb
exact ⟨hx.left, hb⟩
· intro hx
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
have hxab : x = (a, x.snd) := by
ext <;> simp
exact hx.left
rwa [← hxab] at this
· intro ⟨x, ⟨hx, hy⟩⟩
show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
apply And.intro
· simp only [Set.mem_powerset_iff]
rw [hy]
unfold Set.prod
simp only [
Set.mem_singleton_iff,
Set.setOf_subset_setOf,
and_imp,
Prod.forall
]
intro a b ha hb
exact ⟨by rw [ha]; exact hx, hb⟩
· refine ⟨x, ⟨hx, ?_⟩⟩
intro p
apply Iff.intro
· intro hab
rw [hy] at hab
unfold Set.prod at hab
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
· intro ⟨b, ⟨hb, hab⟩⟩
rw [hy]
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
rw [Prod.ext_iff] at hab
simp only at hab
rw [hab.right]
exact ⟨hab.left, hb⟩
/-- ### Exercise 5.5b
With `A`, `B`, and `C` as above, show that `A × B = C`.
-/
theorem exercise_5_5b {A : Set α} (B : Set β)
: Set.prod A B = ⋃₀ {Set.prod ({x} : Set α) B | x ∈ A} := by
suffices Set.prod A B ⊆ ⋃₀ {Set.prod {x} B | x ∈ A} ∧
⋃₀ {Set.prod {x} B | x ∈ A} ⊆ Set.prod A B from
Set.Subset.antisymm_iff.mpr this
apply And.intro
· show ∀ t, t ∈ Set.prod A B → t ∈ ⋃₀ {Set.prod {x} B | x ∈ A}
intro t h
simp only [Set.mem_setOf_eq] at h
unfold Set.sUnion sSup Set.instSupSetSet
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and]
unfold Set.prod at h
simp only [Set.mem_setOf_eq] at h
refine ⟨t.fst, ⟨h.left, ?_⟩⟩
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq, true_and]
exact h.right
· show ∀ t, t ∈ ⋃₀ {Set.prod {x} B | x ∈ A} → t ∈ Set.prod A B
unfold Set.prod
intro t ht
simp only [
Set.mem_singleton_iff,
Set.mem_sUnion,
Set.mem_setOf_eq,
exists_exists_and_eq_and
] at ht
have ⟨a, ⟨h, ⟨ha, hb⟩⟩⟩ := ht
simp only [Set.mem_setOf_eq]
rw [← ha] at h
exact ⟨h, hb⟩
/-- ### Theorem 3D
If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to ` A`.
-/
theorem theorem_3d {A : Set (Set (Set (αα)))} (h : OrderedPair x y ∈ A)
: Sum.inl x ∈ ⋃₀ (⋃₀ A) ∧ Sum.inr y ∈ ⋃₀ (⋃₀ A) := by
have hp : OrderedPair x y ⊆ ⋃₀ A := Chapter_2.exercise_3_3 (OrderedPair x y) h
have hp' : ∀ t, t ∈ {{Sum.inl x}, {Sum.inl x, Sum.inr y}} → t ∈ ⋃₀ A := hp
have hq := hp' {Sum.inl x, Sum.inr y} (by simp)
have hq' := Chapter_2.exercise_3_3 {Sum.inl x, Sum.inr y} hq
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