import Bookshelf.Enderton.Set.Chapter_2 import Bookshelf.Enderton.Set.OrderedPair import Bookshelf.Enderton.Set.Relation import Common.Logic.Basic import Mathlib.Data.Real.Basic import Mathlib.Data.Rel import Mathlib.Init.Algebra.Classes import Mathlib.Order.RelClasses import Mathlib.Tactic.CasesM /-! # Enderton.Set.Chapter_3 Relations and Functions -/ namespace Enderton.Set.Chapter_3 /-- #### Lemma 3B If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`. -/ lemma lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C) : OrderedPair x y ∈ 𝒫 𝒫 C := by have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy exact Set.mem_mem_imp_pair_subset hxs hxys /-- #### Exercise 3.1 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_3_1 {x y z u v w : ℕ} (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 ℕ)) = {{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 3.2a Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`. -/ theorem exercise_3_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 3.2b Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`. -/ theorem exercise_3_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 · rw [Set.Subset.antisymm_iff] 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 3.3 Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`. -/ theorem exercise_3_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 3.5a 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_3_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 3.5b With `A`, `B`, and `C` as above, show that `A × B = ∪ C`. -/ theorem exercise_3_5b {A : Set α} (B : Set β) : Set.prod A B = ⋃₀ {Set.prod ({x} : Set α) B | x ∈ A} := by rw [Set.Subset.antisymm_iff] 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 α))} (h : OrderedPair x y ∈ A) : x ∈ ⋃₀ (⋃₀ A) ∧ y ∈ ⋃₀ (⋃₀ A) := by have hp := Chapter_2.exercise_2_3 (OrderedPair x y) h unfold OrderedPair at hp have hq : {x, y} ∈ ⋃₀ A := hp (by simp) have : {x, y} ⊆ ⋃₀ ⋃₀ A := Chapter_2.exercise_2_3 {x, y} hq exact ⟨this (by simp), this (by simp)⟩ section Relation open Set.Relation /-- #### Exercise 3.6 Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`. -/ theorem exercise_3_6 {A : Set.HRelation α β} : A ⊆ Set.prod (dom A) (ran 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⟩⟩ /-- #### Exercise 3.7 Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`. -/ theorem exercise_3_7 {R : Set.Relation α} : R.fld = ⋃₀ ⋃₀ R.toOrderedPairs := by let img := R.toOrderedPairs rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ x, x ∈ R.fld → x ∈ ⋃₀ ⋃₀ img intro x hx apply Or.elim hx · intro hd unfold Set.Relation.dom Prod.fst at hd simp only [ Set.mem_image, Prod.exists, exists_and_right, exists_eq_right ] at hd have ⟨y, hp⟩ := hd have hm : OrderedPair x y ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by unfold Set.image simp only [Prod.exists, Set.mem_setOf_eq] exact ⟨x, ⟨y, ⟨hp, rfl⟩⟩⟩ unfold OrderedPair at hm have : {x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{x}, {x, y}} hm (by simp) exact (Chapter_2.exercise_2_3 {x} this) (show x ∈ {x} by rfl) · intro hr unfold Set.Relation.ran Prod.snd at hr simp only [Set.mem_image, Prod.exists, exists_eq_right] at hr have ⟨t, ht⟩ := hr have hm : OrderedPair t x ∈ Set.image (fun p => OrderedPair p.1 p.2) R := by simp only [Set.mem_image, Prod.exists] exact ⟨t, ⟨x, ⟨ht, rfl⟩⟩⟩ unfold OrderedPair at hm have : {t, x} ∈ ⋃₀ img := Chapter_2.exercise_2_3 {{t}, {t, x}} hm (show {t, x} ∈ {{t}, {t, x}} by simp) exact Chapter_2.exercise_2_3 {t, x} this (show x ∈ {t, x} by simp) · show ∀ t, t ∈ ⋃₀ ⋃₀ img → t ∈ Set.Relation.fld R intro t ht have ⟨T, hT⟩ : ∃ T ∈ ⋃₀ img, t ∈ T := ht have ⟨T', hT'⟩ : ∃ T' ∈ img, T ∈ T' := hT.left dsimp only at hT' unfold Set.Relation.toOrderedPairs at hT' simp only [Set.mem_image, Prod.exists] at hT' have ⟨x, ⟨y, ⟨p, hp⟩⟩⟩ := hT'.left have hr := hT'.right rw [← hp] at hr unfold OrderedPair at hr simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hr -- Use `exercise_6_6` to prove that if `t = x` then `t ∈ dom R` and if -- `t = y` then `t ∈ ran R`. have hxy_mem : t = x ∨ t = y → t ∈ Set.Relation.fld R := by intro ht have hz : R ⊆ Set.prod (dom R) (ran R) := exercise_3_6 have : (x, y) ∈ Set.prod (dom R) (ran R) := hz p unfold Set.prod at this simp at this apply Or.elim ht · intro ht' rw [← ht'] at this exact Or.inl this.left · intro ht' rw [← ht'] at this exact Or.inr this.right -- Eliminate `T = {x} ∨ T = {x, y}`. apply Or.elim hr · intro hx have := hT.right rw [hx] at this simp only [Set.mem_singleton_iff] at this exact hxy_mem (Or.inl this) · intro hxy have := hT.right rw [hxy] at this simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this exact hxy_mem this /-- #### Exercise 3.8 (i) Show that for any set `𝓐`: ``` dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 } ``` -/ theorem exercise_3_8_i {A : Set (Set.HRelation α β)} : dom (⋃₀ A) = ⋃₀ { dom R | R ∈ A } := by ext x unfold 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 3.8 (ii) Show that for any set `𝓐`: ``` ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 } ``` -/ theorem exercise_3_8_ii {A : Set (Set.HRelation α β)} : ran (⋃₀ A) = ⋃₀ { ran R | R ∈ A } := by ext x unfold 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 3.9 (i) Discuss the result of replacing the union operation by the intersection operation in the preceding problem. ``` dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 } ``` -/ theorem exercise_3_9_i {A : Set (Set.HRelation α β)} : dom (⋂₀ A) ⊆ ⋂₀ { dom R | R ∈ A } := by show ∀ x, x ∈ dom (⋂₀ A) → x ∈ ⋂₀ { dom R | R ∈ A } unfold 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 3.9 (ii) Discuss the result of replacing the union operation by the intersection operation in the preceding problem. ``` ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 } ``` -/ theorem exercise_3_9_ii {A : Set (Set.HRelation α β)} : ran (⋂₀ A) ⊆ ⋂₀ { ran R | R ∈ A } := by show ∀ x, x ∈ ran (⋂₀ A) → x ∈ ⋂₀ { ran R | R ∈ A } unfold 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⟩ /-- #### Theorem 3G (i) Assume that `F` is a one-to-one function. If `x ∈ dom F`, then `F⁻¹(F(x)) = x`. -/ theorem theorem_3g_i {F : Set.HRelation α β} (hF : isOneToOne F) (hx : x ∈ dom F) : ∃! y, (x, y) ∈ F ∧ (y, x) ∈ inv F := by simp only [mem_self_comm_mem_inv, and_self] have ⟨y, hy⟩ := dom_exists hx refine ⟨y, hy, ?_⟩ intro y₁ hy₁ unfold isOneToOne at hF exact (single_valued_eq_unique hF.left hy hy₁).symm /-- #### Theorem 3G (ii) Assume that `F` is a one-to-one function. If `y ∈ ran F`, then `F(F⁻¹(y)) = y`. -/ theorem theorem_3g_ii {F : Set.HRelation α β} (hF : isOneToOne F) (hy : y ∈ ran F) : ∃! x, (x, y) ∈ F ∧ (y, x) ∈ inv F := by simp only [mem_self_comm_mem_inv, and_self] have ⟨x, hx⟩ := ran_exists hy refine ⟨x, hx, ?_⟩ intro x₁ hx₁ unfold isOneToOne at hF exact (single_rooted_eq_unique hF.right hx hx₁).symm /-- #### Theorem 3H Assume that `F` and `G` are functions. Then ``` dom (F ∘ G) = {x ∈ dom G | G(x) ∈ dom F}. ``` -/ theorem theorem_3h_dom {F : Set.HRelation β γ} {G : Set.HRelation α β} (_ : isSingleValued F) (hG : isSingleValued G) : dom (comp F G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F} := by let rhs := {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F } rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ t, t ∈ dom (comp F G) → t ∈ rhs intro t ht simp only [Set.mem_setOf_eq] have ⟨z, hz⟩ := dom_exists ht refine ⟨dom_comp_imp_dom_self ht, ?_⟩ simp only [Set.mem_setOf_eq] at hz have ⟨a, ha⟩ := hz unfold dom simp only [Set.mem_image, Prod.exists, exists_and_right, exists_eq_right] unfold ExistsUnique simp only [and_imp, forall_exists_index] refine ⟨a, ⟨ha.left, z, ha.right⟩, ?_⟩ intro y₁ hy₁ exact fun _ _ => single_valued_eq_unique hG hy₁ ha.left · show ∀ t, t ∈ rhs → t ∈ dom (comp F G) intro t ht simp only [Set.mem_setOf_eq] at ht unfold dom simp only [Set.mem_image, Prod.exists, exists_and_right, exists_eq_right] have ⟨a, ha⟩ := ht.right simp at ha have ⟨b, hb⟩ := dom_exists ha.left.right refine ⟨b, ?_⟩ unfold comp simp only [Set.mem_setOf_eq] exact ⟨a, ha.left.left, hb⟩ /-- #### Theorem 3J (a) Assume that `F : A → B`, and that `A` is nonempty. There exists a function `G : B → A` (a "left inverse") such that `G ∘ F` is the identity function on `A` **iff** `F` is one-to-one. -/ theorem theorem_3j_a {F : Set.HRelation α β} {A : Set α} {B : Set β} (hF : mapsInto F A B) (hA : Set.Nonempty A) : (∃ G : Set.HRelation β α, mapsInto G B A ∧ (comp G F = { p | p.1 ∈ A ∧ p.1 = p.2 })) ↔ isOneToOne F := by apply Iff.intro · intro ⟨G, hG⟩ refine ⟨hF.is_func, ?_⟩ intro y hy have ⟨x₁, hx₁⟩ := ran_exists hy refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩ intro x₂ hx₂ have hG' : y ∈ dom G := by rw [hG.left.dom_eq] exact hF.ran_ss hy have ⟨z, hz⟩ := dom_exists hG' have := hG.right unfold comp at this rw [Set.ext_iff] at this have h₁ := (this (x₁, z)).mp ⟨y, hx₁, hz⟩ have h₂ := (this (x₂, z)).mp ⟨y, hx₂.right, hz⟩ simp only [Set.mem_setOf_eq] at h₁ h₂ rw [h₁.right, h₂.right] · sorry /-- #### Theorem 3J (b) Assume that `F : A → B`, and that `A` is nonempty. There exists a function `H : B → A` (a "right inverse") such that `F ∘ H` is the identity function on `B` **iff** `F` maps `A` onto `B`. -/ theorem theorem_3j_b {F : Set.HRelation α β} {A : Set α} {B : Set β} (hF : mapsInto F A B) (hA : Set.Nonempty A) : (∃ H : Set.HRelation β α, mapsInto H B A ∧ (comp F H = { p | p.1 ∈ B ∧ p.1 = p.2 })) ↔ mapsOnto F 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.HRelation α β} {𝓐 : Set (Set α)} : image F (⋃₀ 𝓐) = ⋃₀ { image F A | A ∈ 𝓐 } := by rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ v, v ∈ image F (⋃₀ 𝓐) → v ∈ ⋃₀ { image F 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 ∈ image F A unfold image simp only [Set.mem_setOf_eq] exact ⟨u, hA.right, hu.right⟩ · show ∀ v, v ∈ ⋃₀ {x | ∃ A, A ∈ 𝓐 ∧ image F A = x} → v ∈ image F (⋃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.HRelation α β} {𝓐 : Set (Set α)} : image F (⋂₀ 𝓐) ⊆ ⋂₀ { image F A | A ∈ 𝓐} := by show ∀ v, v ∈ image F (⋂₀ 𝓐) → v ∈ ⋂₀ { image F 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.HRelation α β} {𝓐 : Set (Set α)} (hF : isSingleRooted F) (h𝓐 : Set.Nonempty 𝓐) : image F (⋂₀ 𝓐) = ⋂₀ { image F 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.HRelation α β} {A B : Set α} : image F A \ image F B ⊆ image F (A \ B) := by show ∀ v, v ∈ image F A \ image F B → v ∈ image F (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.HRelation α β} {A B : Set α} (hF : isSingleRooted F) : image F A \ image F B = image F (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 ∈ image F A := by unfold image simp only [Set.mem_setOf_eq] exact ⟨u, hu.left.left, hu.right⟩ have hv₂ : v ∉ image F B := by intro nv unfold image at nv simp only [Set.mem_setOf_eq] at nv have ⟨u₁, hu₁⟩ := nv have := single_rooted_eq_unique hF hu.right hu₁.right rw [← this] 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.HRelation β α} {𝓐 : Set (Set α)} : image (inv G) (⋃₀ 𝓐) = ⋃₀ {image (inv G) A | A ∈ 𝓐} := theorem_3k_a theorem corollary_3l_ii {G : Set.HRelation β α} {𝓐 : Set (Set α)} (hG : isSingleValued G) (h𝓐 : Set.Nonempty 𝓐) : image (inv G) (⋂₀ 𝓐) = ⋂₀ {image (inv G) A | A ∈ 𝓐} := by have hG' : isSingleRooted (inv G) := single_valued_self_iff_single_rooted_inv.mp hG exact theorem_3k_b_ii hG' h𝓐 theorem corollary_3l_iii {G : Set.HRelation β α} {A B : Set α} (hG : isSingleValued G) : image (inv G) (A \ B) = image (inv G) A \ image (inv G) B := by have hG' : isSingleRooted (inv G) := single_valued_self_iff_single_rooted_inv.mp hG exact (theorem_3k_c_ii hG').symm /-- #### Exercise 3.12 Assume that `f` and `g` are functions and show that ``` f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f) f(x) = g(x). ``` -/ theorem exercise_3_12 {f g : Set.HRelation α β} (hf : isSingleValued f) (_ : isSingleValued g) : f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f, ∃! y : β, (x, y) ∈ f ∧ (x, y) ∈ g) := by apply Iff.intro · intro h apply And.intro · show ∀ x, x ∈ dom f → x ∈ dom g intro x hx have ⟨y, hy⟩ := dom_exists hx exact mem_pair_imp_fst_mem_dom (h hy) · intro x hx have ⟨y, hy⟩ := dom_exists hx refine ⟨y, ⟨hy, h hy⟩, ?_⟩ intro y₁ hy₁ exact single_valued_eq_unique hf hy₁.left hy · intro ⟨_, hx⟩ show ∀ p, p ∈ f → p ∈ g intro (x, y) hp have ⟨y₁, hy₁⟩ := hx x (mem_pair_imp_fst_mem_dom hp) rw [single_valued_eq_unique hf hp hy₁.left.left] exact hy₁.left.right /-- #### Exercise 3.13 Assume that `f` and `g` are functions with `f ⊆ g` and `dom g ⊆ dom f`. Show that `f = g`. -/ theorem exercise_3_13 {f g : Set.HRelation α β} (hf : isSingleValued f) (hg : isSingleValued g) (h : f ⊆ g) (h₁ : dom g ⊆ dom f) : f = g := by have h₂ := (exercise_3_12 hf hg).mp h have hfg := Set.Subset.antisymm_iff.mpr ⟨h₁, h₂.left⟩ ext p have (a, b) := p apply Iff.intro · intro hp have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp) rw [single_valued_eq_unique hf hp hx.left.left] exact hx.left.right · intro hp rw [← hfg] at h₂ have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp) rw [single_valued_eq_unique hg hp hx.left.right] exact hx.left.left /-- #### Exercise 3.14 (a) Assume that `f` and `g` are functions. Show that `f ∩ g` is a function. -/ theorem exercise_3_14_a {f g : Set.HRelation α β} (hf : isSingleValued f) (_ : isSingleValued g) : isSingleValued (f ∩ g) := single_valued_subset hf (Set.inter_subset_left f g) /-- #### Exercise 3.14 (b) Assume that `f` and `g` are functions. Show that `f ∪ g` is a function **iff** `f(x) = g(x)` for every `x` in `(dom f) ∩ (dom g)`. -/ theorem exercise_3_14_b {f g : Set.HRelation α β} (hf : isSingleValued f) (hg : isSingleValued g) : isSingleValued (f ∪ g) ↔ (∀ x ∈ dom f ∩ dom g, ∃! y, (x, y) ∈ f ∧ (x, y) ∈ g) := by apply Iff.intro · intro h x hx have ⟨y₁, hy₁⟩ := hf x hx.left have ⟨y₂, hy₂⟩ := hg x hx.right have : y₁ = y₂ := single_valued_eq_unique h (Or.inl hy₁.left.right) (Or.inr hy₂.left.right) rw [← this] at hy₂ refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩ intro y₃ hfy₃ exact single_valued_eq_unique hf hfy₃.left hy₁.left.right · intro h x hx by_cases hfx : x ∈ dom f <;> by_cases hgx : x ∈ dom g · -- `x ∈ dom f ∧ x ∈ dom g` have ⟨y₁, hy₁⟩ := hf x hfx have ⟨y₂, hy₂⟩ := hg x hgx refine ⟨y₁, ⟨?_, Or.inl hy₁.left.right⟩, ?_⟩ · unfold ran simp only [Set.mem_image, Set.mem_union, Prod.exists, exists_eq_right] exact ⟨x, Or.inl hy₁.left.right⟩ · intro y₃ hy₃ apply Or.elim hy₃.right · intro hxy exact single_valued_eq_unique hf hxy hy₁.left.right · refine fun hxy => single_valued_eq_unique hg hxy ?_ have : y₁ = y₂ := by have ⟨z, ⟨hz, _⟩⟩ := h x ⟨hfx, hgx⟩ rw [ single_valued_eq_unique hf hy₁.left.right hz.left, single_valued_eq_unique hg hy₂.left.right hz.right ] rw [this] exact hy₂.left.right · -- `x ∈ dom f ∧ x ∉ dom g` have ⟨y, hy⟩ := dom_exists hfx have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_left f g) hy refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩ intro y₁ hy₁ apply Or.elim hy₁.right · intro hx' exact single_valued_eq_unique hf hx' hy · intro hx' exact absurd (mem_pair_imp_fst_mem_dom hx') hgx · -- `x ∉ dom f ∧ x ∈ dom g` have ⟨y, hy⟩ := dom_exists hgx have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_right f g) hy refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩ intro y₁ hy₁ apply Or.elim hy₁.right · intro hx' exact absurd (mem_pair_imp_fst_mem_dom hx') hfx · intro hx' exact single_valued_eq_unique hg hx' hy · -- `x ∉ dom f ∧ x ∉ dom g` exfalso unfold dom at hx simp only [ Set.mem_image, Set.mem_union, Prod.exists, exists_and_right, exists_eq_right ] at hx have ⟨_, hy⟩ := hx apply Or.elim hy · intro hz exact absurd (mem_pair_imp_fst_mem_dom hz) hfx · intro hz exact absurd (mem_pair_imp_fst_mem_dom hz) hgx /-- #### Exercise 3.15 Let `𝓐` be a set of functions such that for any `f` and `g` in `𝓐`, either `f ⊆ g` or `g ⊆ f`. Show that `⋃ 𝓐` is a function. -/ theorem exercise_3_15 {𝓐 : Set (Set.HRelation α β)} (h𝓐 : ∀ F ∈ 𝓐, isSingleValued F) (h : ∀ F, ∀ G, F ∈ 𝓐 → G ∈ 𝓐 → F ⊆ G ∨ G ⊆ F) : isSingleValued (⋃₀ 𝓐) := by intro x hx have ⟨y₁, hy₁⟩ := dom_exists hx refine ⟨y₁, ⟨mem_pair_imp_snd_mem_ran hy₁, hy₁⟩, ?_⟩ intro y₂ hy₂ have ⟨f, hf⟩ : ∃ f : Set.HRelation α β, f ∈ 𝓐 ∧ (x, y₁) ∈ f := hy₁ have ⟨g, hg⟩ : ∃ g : Set.HRelation α β, g ∈ 𝓐 ∧ (x, y₂) ∈ g := hy₂.right apply Or.elim (h f g hf.left hg.left) · intro hf' have := hf' hf.right exact single_valued_eq_unique (h𝓐 g hg.left) hg.right this · intro hg' have := hg' hg.right exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right /-! #### Exercise 3.17 Show that the composition of two single-rooted sets is again single-rooted. Conclude that the composition of two one-to-one functions is again one-to-one. -/ theorem exercise_3_17_i {F : Set.HRelation β γ} {G : Set.HRelation α β} (hF : isSingleRooted F) (hG : isSingleRooted G) : isSingleRooted (comp F G):= by intro v hv have ⟨u₁, hu₁⟩ := ran_exists hv have hu₁' := hu₁ unfold comp at hu₁' simp only [Set.mem_setOf_eq] at hu₁' have ⟨t₁, ht₁⟩ := hu₁' unfold ExistsUnique refine ⟨u₁, ⟨mem_pair_imp_fst_mem_dom hu₁, hu₁⟩, ?_⟩ intro u₂ hu₂ have hu₂' := hu₂ unfold comp at hu₂' simp only [Set.mem_setOf_eq] at hu₂' have ⟨_, ⟨t₂, ht₂⟩⟩ := hu₂' have ht : t₁ = t₂ := single_rooted_eq_unique hF ht₁.right ht₂.right rw [ht] at ht₁ exact single_rooted_eq_unique hG ht₂.left ht₁.left theorem exercise_3_17_ii {F : Set.HRelation β γ} {G : Set.HRelation α β} (hF : isOneToOne F) (hG : isOneToOne G) : isOneToOne (comp F G) := And.intro (single_valued_comp_is_single_valued hF.left hG.left) (exercise_3_17_i hF.right hG.right) /-! #### Exercise 3.18 Let `R` be the set ``` {⟨0, 1⟩, ⟨0, 2⟩, ⟨0, 3⟩, ⟨1, 2⟩, ⟨1, 3⟩, ⟨2, 3⟩} ``` Evaluate the following: `R ∘ R`, `R ↾ {1}`, `R⁻¹ ↾ {1}`, `R⟦{1}⟧`, and `R⁻¹⟦{1}⟧`. -/ section Exercise_3_18 variable {R : Set.Relation ℕ} variable (hR : R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}) theorem exercise_3_18_i : comp R R = {(0, 2), (0, 3), (1, 3)} := by rw [hR] unfold comp simp only [Set.mem_singleton_iff, Set.mem_insert_iff, or_self, Prod.mk.injEq] ext x have (a, b) := x apply Iff.intro · simp only [Set.mem_setOf_eq, Set.mem_singleton_iff, Set.mem_insert_iff] intro ⟨t, ht₁, ht₂⟩ casesm* _ ∨ _ all_goals case _ hl hr => first | {rw [hl.right] at hr; simp at hr} | {rw [hl.left] at hr; simp at hr} | {rw [hl.left, hr.right]; simp} · simp only [ Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq, Set.mem_setOf_eq ] intro h casesm* _ ∨ _ · case _ h => refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩ iterate 3 right left exact ⟨rfl, h.right⟩ · case _ h => refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩ iterate 4 right left exact ⟨rfl, h.right⟩ · case _ h => refine ⟨2, ?_, ?_⟩ · iterate 3 right left exact ⟨h.left, rfl⟩ · iterate 5 right exact ⟨rfl, h.right⟩ theorem exercise_3_18_ii : restriction R {1} = {(1, 2), (1, 3)} := by rw [hR] unfold restriction ext p have (a, b) := p simp only [ Set.mem_singleton_iff, Set.mem_insert_iff, Set.mem_setOf_eq, or_self ] apply Iff.intro · intro ⟨hp, ha⟩ rw [ha] simp only [Prod.mk.injEq, true_and] casesm* _ ∨ _ all_goals case _ h => first | {rw [ha] at h; simp at h} | {simp only [Prod.mk.injEq] at h; left; exact h.right} | {simp only [Prod.mk.injEq] at h; right; exact h.right} · intro h apply Or.elim h · intro hab simp only [Prod.mk.injEq] at hab refine ⟨?_, hab.left⟩ iterate 3 right left rw [hab.left, hab.right] · intro hab simp only [Prod.mk.injEq] at hab refine ⟨?_, hab.left⟩ iterate 4 right left rw [hab.left, hab.right] theorem exercise_3_18_iii : restriction (inv R) {1} = {(1, 0)} := by rw [hR] unfold inv restriction ext p have (a, b) := p simp only [ Set.mem_singleton_iff, Set.mem_insert_iff, or_self, exists_eq_or_imp, exists_eq_left, Set.mem_setOf_eq, Prod.mk.injEq ] apply Iff.intro · intro ⟨hb, ha⟩ casesm* _ ∨ _ all_goals case _ hr => first | exact ⟨ha, hr.right.symm⟩ | rw [ha] at hr; simp at hr · intro ⟨ha, hb⟩ rw [ha, hb] simp theorem exercise_3_18_iv : image R {1} = {2, 3} := by rw [hR] unfold image ext y simp theorem exercise_3_18_v : image (inv R) {1} = {0} := by rw [hR] unfold inv image ext y simp end Exercise_3_18 /-! #### Exercise 3.19 Let ``` A = {⟨∅, {∅, {∅}}⟩, ⟨{∅}, ∅⟩}. ``` Evaluate each of the following: `A(∅)`, `A⟦∅⟧`, `A⟦{∅}⟧`, `A⟦{∅, {∅}}⟧`, `A⁻¹`, `A ∘ A`, `A ↾ ∅`, `A ↾ {∅, {∅}}`, `⋃ ⋃ A`. -/ section Exercise_3_19 variable {A : Set.Relation (Set (Set (Set α)))} variable (hA : A = {(∅, {∅, {∅}}), ({∅}, ∅)}) theorem exercise_3_19_i : (∅, {∅, {∅}}) ∈ A := by rw [hA] simp theorem exercise_3_19_ii : image A ∅ = ∅ := by unfold image simp theorem exercise_3_19_iii : image A {∅} = {{∅, {∅}}} := by unfold image rw [hA] ext x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, exists_eq_left, true_and ] apply Iff.intro · intro hx simp at hx apply Or.elim hx · simp · intro ⟨h, _⟩ exfalso rw [Set.ext_iff] at h have := h ∅ simp at this · intro hx rw [hx] simp theorem exercise_3_19_iv : image A {∅, {∅}} = {{∅, {∅}}, ∅} := by unfold image rw [hA] ext x simp only [ Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq, exists_eq_or_imp, true_and, exists_eq_left, Set.singleton_ne_empty, false_and, false_or, Set.mem_setOf_eq ] apply Iff.intro · intro h apply Or.elim h · intro hx₁ apply Or.elim hx₁ · intro hx₂; left ; exact hx₂ · intro hx₂; right; exact hx₂.right · intro hx₂ right exact hx₂ · intro h apply Or.elim h · intro hx₁; iterate 2 left exact hx₁ · intro hx₁; right; exact hx₁ theorem exercise_3_19_v : inv A = {({∅, {∅}}, ∅), (∅, {∅})} := by unfold inv rw [hA] ext x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, exists_eq_or_imp, exists_eq_left, Set.mem_setOf_eq ] apply Iff.intro · intro hx apply Or.elim hx · intro hx₁; left ; rw [← hx₁] · intro hx₁; right; rw [← hx₁] · intro hx apply Or.elim hx · intro hx₁; left ; rw [← hx₁] · intro hx₁; right; rw [← hx₁] theorem exercise_3_19_vi : comp A A = {({∅}, {∅, {∅}})} := by unfold comp rw [hA] ext x have (a, b) := x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, Set.mem_setOf_eq ] apply Iff.intro · intro ⟨t, ht₁, ht₂⟩ casesm* _ ∨ _ all_goals case _ hl hr => first | { rw [hl.right] at hr have := hr.left rw [Set.ext_iff] at this simp at this } | exact ⟨hl.left, hr.right⟩ · intro ⟨ha, hb⟩ refine ⟨∅, ?_, ?_⟩ · right; rw [ha]; simp · left ; rw [hb]; simp theorem exercise_3_19_vii : restriction A ∅ = ∅ := by unfold restriction rw [hA] simp theorem exercise_3_19_viii : restriction A {∅} = {(∅, {∅, {∅}})} := by unfold restriction rw [hA] ext x have (a, b) := x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, Set.mem_setOf_eq ] apply Iff.intro · intro ⟨h, ha⟩ apply Or.elim h · simp · intro ⟨ha', _⟩ exfalso rw [ha', Set.ext_iff] at ha simp at ha · intro ⟨ha, hb⟩ rw [ha, hb] simp theorem exercise_3_19_ix : restriction A {∅, {∅}} = A := by unfold restriction rw [hA] ext x have (a, b) := x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_insert_iff, Set.mem_setOf_eq ] apply Iff.intro · intro ⟨h₁, h₂⟩ casesm* _ ∨ _ · case _ hl _ => left ; exact hl · case _ hl _ => left ; exact hl · case _ hl _ => right; exact hl · case _ hl _ => right; exact hl · intro h₁ apply Or.elim h₁ <;> · intro ⟨ha, hb⟩ rw [ha, hb] simp theorem exercise_3_19_x : ⋃₀ ⋃₀ A.toOrderedPairs = {∅, {∅}, {∅, {∅}}} := by unfold toOrderedPairs OrderedPair Set.sUnion sSup Set.instSupSetSet rw [hA] ext x simp only [ Set.mem_singleton_iff, Prod.mk.injEq, Set.mem_image, Set.mem_insert_iff, exists_eq_or_imp, exists_eq_left, Set.singleton_ne_empty, Set.mem_setOf_eq ] apply Iff.intro · intro ⟨a, ⟨t, ht₁, ht₂⟩, hx⟩ apply Or.elim ht₁ · intro ht rw [← ht] at ht₂ simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at ht₂ apply Or.elim ht₂ · intro ha rw [ha] at hx simp only [Set.mem_singleton_iff] at hx left exact hx · intro ha rw [ha] at hx simp at hx apply Or.elim hx <;> · intro hx'; rw [hx']; simp · intro ht rw [← ht] at ht₂ simp only [ Set.mem_singleton_iff, Set.singleton_ne_empty, Set.mem_insert_iff ] at ht₂ apply Or.elim ht₂ · intro ha rw [ha] at hx simp only [Set.mem_singleton_iff] at hx rw [hx] simp · intro ha rw [ha] at hx simp only [ Set.mem_singleton_iff, Set.singleton_ne_empty, Set.mem_insert_iff ] at hx apply Or.elim hx <;> · intro hx'; rw [hx']; simp · intro hx apply Or.elim hx · intro hx₁ rw [hx₁] refine ⟨{{∅}, ∅}, ⟨{{{∅}}, {{∅}, ∅}}, ?_⟩, ?_⟩ <;> simp · intro hx₁ apply Or.elim hx₁ · intro hx₂ rw [hx₂] refine ⟨{{∅}, ∅}, ⟨{{{∅}}, {{∅}, ∅}}, ?_⟩, ?_⟩ <;> simp · intro hx₂ rw [hx₂] refine ⟨{∅, {∅, {∅}}}, ⟨{{∅}, {∅, {∅, {∅}}}}, ?_⟩, ?_⟩ <;> simp end Exercise_3_19 /-- #### Exercise 3.20 Show that `F ↾ A = F ∩ (A × ran F)`. -/ theorem exercise_3_20 {F : Set.HRelation α β} {A : Set α} : restriction F A = F ∩ (Set.prod A (ran F)) := by calc restriction F A _ = {p | p ∈ F ∧ p.fst ∈ A} := rfl _ = {p | p ∈ F ∧ p.fst ∈ A ∧ p.snd ∈ ran F} := by ext x have (a, b) := x simp only [ Set.mem_setOf_eq, Set.sep_and, Set.mem_inter_iff, iff_self_and, and_imp ] intro hF _ exact ⟨hF, mem_pair_imp_snd_mem_ran hF⟩ _ = {p | p ∈ F} ∩ {p | p.fst ∈ A ∧ p.snd ∈ ran F} := rfl _ = F ∩ {p | p.fst ∈ A ∧ p.snd ∈ ran F} := rfl _ = F ∩ (Set.prod A (ran F)) := rfl /-- #### Exercise 3.22 (a) Show that the following is correct for any sets. ``` A ⊆ B → F⟦A⟧ ⊆ F⟦B⟧ ``` -/ theorem exercise_3_22_a {A B : Set α} {F : Set.HRelation α β} (h : A ⊆ B) : image F A ⊆ image F B := by show ∀ x, x ∈ image F A → x ∈ image F B intro x hx have ⟨u, hu⟩ := hx have := h hu.left exact ⟨u, this, hu.right⟩ /-- #### Exercise 3.22 (b) Show that the following is correct for any sets. ``` (F ∘ G)⟦A⟧ = F⟦G⟦A⟧⟧ ``` -/ theorem exercise_3_22_b {A B : Set α} {F : Set.HRelation α β} : image (comp F G) A = image F (image G A) := by calc image (comp F G) A _ = { v | ∃ u ∈ A, (u, v) ∈ comp F G } := rfl _ = { v | ∃ u ∈ A, ∃ a, (u, a) ∈ G ∧ (a, v) ∈ F } := rfl _ = { v | ∃ a, ∃ u ∈ A, (u, a) ∈ G ∧ (a, v) ∈ F } := by ext p apply Iff.intro · intro ⟨u, hu, a, ha⟩ exact ⟨a, u, hu, ha⟩ · intro ⟨a, u, hu, ha⟩ exact ⟨u, hu, a, ha⟩ _ = { v | ∃ a, (∃ u ∈ A, (u, a) ∈ G) ∧ (a, v) ∈ F } := by ext p apply Iff.intro · intro ⟨a, u, h⟩ exact ⟨a, ⟨u, h.left, h.right.left⟩, h.right.right⟩ · intro ⟨a, ⟨u, hu⟩, ha⟩ exact ⟨a, u, hu.left, hu.right, ha⟩ _ = { v | ∃ a ∈ { w | ∃ u ∈ A, (u, w) ∈ G }, (a, v) ∈ F } := rfl _ = { v | ∃ a ∈ image G A, (a, v) ∈ F } := rfl _ = image F (image G A) := rfl /-- #### Exercise 3.22 (c) Show that the following is correct for any sets. ``` Q ↾ (A ∪ B) = (Q ↾ A) ∪ (Q ↾ B) ``` -/ theorem exercise_3_22_c {A B : Set α} {Q : Set.Relation α} : restriction Q (A ∪ B) = (restriction Q A) ∪ (restriction Q B) := by calc restriction Q (A ∪ B) _ = { p | p ∈ Q ∧ p.1 ∈ A ∪ B } := rfl _ = { p | p ∈ Q ∧ (p.1 ∈ A ∨ p.1 ∈ B) } := rfl _ = { p | (p ∈ Q ∧ p.1 ∈ A) ∨ (p ∈ Q ∧ p.1 ∈ B) } := by ext p simp only [Set.sep_or, Set.mem_union, Set.mem_setOf_eq] _ = { p | p ∈ Q ∧ p.1 ∈ A} ∪ { p | p ∈ Q ∧ p.1 ∈ B } := rfl _ = (restriction Q A) ∪ (restriction Q B) := rfl /-- #### Exercise 3.23 (i) Let `I` be the identity function on the set `A`. Show that for any sets `B` and `C`, `B ∘ I = B ↾ A`. -/ theorem exercise_3_23_i {A : Set α} {B : Set.HRelation α β} {I : Set.Relation α} (hI : I = { p | p.1 ∈ A ∧ p.1 = p.2 }) : comp B I = restriction B A := by rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ p, p ∈ comp B I → p ∈ restriction B A intro (x, y) hp have ⟨t, ht⟩ := hp rw [hI] at ht simp only [Set.mem_setOf_eq] at ht show (x, y) ∈ B ∧ x ∈ A rw [← ht.left.right] at ht exact ⟨ht.right, ht.left.left⟩ · show ∀ p, p ∈ restriction B A → p ∈ comp B I unfold restriction comp rw [hI] intro (x, y) hp refine ⟨x, ⟨hp.right, rfl⟩, hp.left⟩ /-- #### Exercise 3.23 (ii) Let `I` be the identity function on the set `A`. Show that for any sets `B` and `C`, `I⟦C⟧ = A ∩ C`. -/ theorem exercise_3_23_ii {A C : Set α} {I : Set.Relation α} (hI : I = { p | p.1 ∈ A ∧ p.1 = p.2 }) : image I C = A ∩ C := by calc image I C _ = { v | ∃ u ∈ C, (u, v) ∈ I } := rfl _ = { v | ∃ u ∈ C, u ∈ A ∧ u = v } := by ext v apply Iff.intro · intro ⟨u, h₁, h₂⟩ rw [hI] at h₂ exact ⟨u, h₁, h₂⟩ · intro ⟨u, h₁, h₂⟩ refine ⟨u, h₁, ?_⟩ · rw [hI] exact h₂ _ = { v | v ∈ C ∧ v ∈ A } := by ext v simp only [Set.mem_setOf_eq, Set.sep_mem_eq, Set.mem_inter_iff] apply Iff.intro · intro ⟨u, hC, hA, hv⟩ rw [← hv] exact ⟨hC, hA⟩ · intro ⟨hC, hA⟩ exact ⟨v, hC, hA, rfl⟩ _ = C ∩ A := rfl _ = A ∩ C := Set.inter_comm C A /-- #### Exercise 3.24 Show that for a function `F`, `F⁻¹⟦A⟧ = { x ∈ dom F | F(x) ∈ A }`. -/ theorem exercise_3_24 {F : Set.HRelation α β} {A : Set β} (hF : isSingleValued F) : image (inv F) A = { x ∈ dom F | ∃! y : β, (x, y) ∈ F ∧ y ∈ A } := by calc image (inv F) A _ = { x | ∃ y ∈ A, (y, x) ∈ inv F } := rfl _ = { x | ∃ y ∈ A, (x, y) ∈ F } := by simp only [mem_self_comm_mem_inv] _ = { x | x ∈ dom F ∧ (∃ y : β, (x, y) ∈ F ∧ y ∈ A) } := by ext x apply Iff.intro · intro ⟨y, hy, hyx⟩ exact ⟨mem_pair_imp_fst_mem_dom hyx, y, hyx, hy⟩ · intro ⟨_, y, hxy, hy⟩ exact ⟨y, hy, hxy⟩ _ = { x ∈ dom F | ∃ y : β, (x, y) ∈ F ∧ y ∈ A } := rfl _ = { x ∈ dom F | ∃! y : β, (x, y) ∈ F ∧ y ∈ A } := by ext x simp only [Set.mem_setOf_eq, and_congr_right_iff] intro _ apply Iff.intro · intro ⟨y, hy⟩ refine ⟨y, hy, ?_⟩ intro y₁ hy₁ exact single_valued_eq_unique hF hy₁.left hy.left · intro ⟨y, hy⟩ exact ⟨y, hy.left⟩ /-- #### Exercise 3.25 (b) Show that the result of part (a) holds for any function `G`, not necessarily one-to-one. -/ theorem exercise_3_25_b {G : Set.HRelation α β} (hG : isSingleValued G) : comp G (inv G) = { p | p.1 ∈ ran G ∧ p.1 = p.2 } := by ext p have (x, y) := p apply Iff.intro · unfold comp inv intro h simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq] at h have ⟨t, ⟨a, b, ⟨hab, hb, ha⟩⟩, ht⟩ := h rw [hb, ha] at hab exact ⟨mem_pair_imp_snd_mem_ran hab, single_valued_eq_unique hG hab ht⟩ · intro h simp only [Set.mem_setOf_eq] at h unfold comp inv simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq] have ⟨t, ht⟩ := ran_exists h.left exact ⟨t, ⟨t, x, ht, rfl, rfl⟩, by rwa [← h.right]⟩ /-- #### Exercise 3.25 (a) Assume that `G` is a one-to-one function. Show that `G ∘ G⁻¹` is the identity function on `ran G`. -/ theorem exercise_3_25_a {G : Set.HRelation α β} (hG : isOneToOne G) : comp G (inv G) = { p | p.1 ∈ ran G ∧ p.1 = p.2 } := exercise_3_25_b hG.left /-- #### Exercise 3.27 Show that `dom (F ∘ G) = G⁻¹⟦dom F⟧` for any sets `F` and `G`. (`F` and `G` need not be functions.) -/ theorem exercise_3_27 {F : Set.HRelation β γ} {G : Set.HRelation α β} : dom (comp F G) = image (inv G) (dom F) := by rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ x, x ∈ dom (comp F G) → x ∈ image (inv G) (dom F) intro x hx have ⟨y, t, ht⟩ := dom_exists hx have htF : t ∈ dom F := mem_pair_imp_fst_mem_dom ht.right unfold image inv simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq] exact ⟨t, htF, x, t, ht.left, rfl, rfl⟩ · show ∀ x, x ∈ image (inv G) (dom F) → x ∈ dom (comp F G) intro x hx unfold image at hx simp only [mem_self_comm_mem_inv, Set.mem_setOf_eq] at hx have ⟨u, hu⟩ := hx have ⟨t, ht⟩ := dom_exists hu.left unfold dom comp simp only [ Set.mem_image, Set.mem_setOf_eq, Prod.exists, exists_and_right, exists_eq_right ] exact ⟨t, u, hu.right, ht⟩ /-- #### Exercise 3.28 Assume that `f` is a one-to-one function from `A` into `B`, and that `G` is the function with `dom G = 𝒫 A` defined by the equation `G(X) = f⟦X⟧`. Show that `G` maps `𝒫 A` one-to-one into `𝒫 B`. -/ theorem exercise_3_28 {A : Set α} {B : Set β} {f : Set.HRelation α β} {G : Set.HRelation (Set α) (Set β)} (hf : isOneToOne f ∧ mapsInto f A B) (hG : G = { p | p.1 ∈ 𝒫 A ∧ p.2 = image f p.1 }) : isOneToOne G ∧ mapsInto G (𝒫 A) (𝒫 B) := by have dG : dom G = 𝒫 A := by rw [hG] ext p unfold dom Prod.fst simp have hG₁ : isSingleValued G := by intro x hx have ⟨y, hy⟩ := dom_exists hx refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hy, hy⟩, ?_⟩ intro y₁ hy₁ rw [hG, Set.mem_setOf_eq] at hy conv at hy₁ => rhs; rw [hG, Set.mem_setOf_eq] simp only at * rw [hy.right, hy₁.right.right] apply And.intro · show isOneToOne G refine ⟨hG₁, ?_⟩ intro y hy have ⟨X₁, hX₁⟩ := ran_exists hy refine ⟨X₁, ⟨mem_pair_imp_fst_mem_dom hX₁, hX₁⟩, ?_⟩ intro X₂ hX₂ have hX₁' : y = image f X₁ := by rw [hG] at hX₁ simp only [Set.mem_powerset_iff, Set.mem_setOf_eq] at hX₁ exact hX₁.right have hX₂' : y = image f X₂ := by have := hX₂.right rw [hG] at this simp only [Set.mem_powerset_iff, Set.mem_setOf_eq] at this exact this.right ext t apply Iff.intro · intro ht rw [dG] at hX₂ simp only [Set.mem_powerset_iff] at hX₂ have ht' := hX₂.left ht rw [← hf.right.dom_eq] at ht' have ⟨ft, hft⟩ := dom_exists ht' have hft' : ft ∈ image f X₂ := ⟨t, ht, hft⟩ rw [← hX₂', hX₁'] at hft' have ⟨t₁, ht₁⟩ := hft' rw [single_rooted_eq_unique hf.left.right hft ht₁.right] exact ht₁.left · intro ht have hX₁sub := mem_pair_imp_fst_mem_dom hX₁ rw [dG] at hX₁sub simp only [Set.mem_powerset_iff] at hX₁sub have ht' := hX₁sub ht rw [← hf.right.dom_eq] at ht' have ⟨ft, hft⟩ := dom_exists ht' have hft' : ft ∈ image f X₁ := ⟨t, ht, hft⟩ rw [← hX₁', hX₂'] at hft' have ⟨t₁, ht₁⟩ := hft' rw [single_rooted_eq_unique hf.left.right hft ht₁.right] exact ht₁.left · show mapsInto G (𝒫 A) (𝒫 B) refine ⟨hG₁, dG, ?_⟩ show ∀ x, x ∈ ran G → x ∈ 𝒫 B intro x hx rw [hG] at hx unfold ran Prod.snd at hx simp only [ Set.mem_powerset_iff, Set.mem_image, Set.mem_setOf_eq, Prod.exists, exists_eq_right ] at hx have ⟨a, ha⟩ := hx rw [ha.right] show ∀ y, y ∈ image f a → y ∈ B intro y ⟨b, hb⟩ have hz := mem_pair_imp_snd_mem_ran hb.right exact hf.right.ran_ss hz /-- #### Exercise 3.29 Assume that `f : A → B` and define a function `G : B → 𝒫 A` by ``` G(b) = {x ∈ A | f(x) = b} ``` Show that if `f` maps `A` *onto* `B`, then `G` is one-to-one. Does the converse hold? -/ theorem exercise_3_29 {f : Set.HRelation α β} {G : Set.HRelation β (Set α)} {A : Set α} {B : Set β} (hf : mapsOnto f A B) (hG : mapsInto G B (𝒫 A) ∧ G = { p | p.1 ∈ B ∧ p.2 = {x ∈ A | (x, p.1) ∈ f} }) : isOneToOne G := by unfold isOneToOne refine ⟨hG.left.is_func, ?_⟩ intro y hy have ⟨x₁, hx₁⟩ := ran_exists hy refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩ intro x₂ hx₂ have hG₁ : (x₁, {x ∈ A | (x, x₁) ∈ f}) ∈ G := by rw [hG.right, ← hG.left.dom_eq] simp only [Set.mem_setOf_eq, and_true] exact mem_pair_imp_fst_mem_dom hx₁ have hG₂ : (x₂, {x ∈ A | (x, x₂) ∈ f}) ∈ G := by rw [hG.right, ← hG.left.dom_eq] simp only [Set.mem_setOf_eq, and_true] exact hx₂.left have heq : {x ∈ A | (x, x₁) ∈ f} = {x ∈ A | (x, x₂) ∈ f} := by have h₁ := single_valued_eq_unique hG.left.is_func hx₁ hG₁ have h₂ := single_valued_eq_unique hG.left.is_func hx₂.right hG₂ rw [← h₁, ← h₂] rw [hG.right, ← hf.ran_eq] at hG₁ simp only [Set.mem_setOf_eq, and_true] at hG₁ have ⟨t, ht⟩ := ran_exists hG₁ have : t ∈ {x ∈ A | (x, x₁) ∈ f} := by refine ⟨?_, ht⟩ rw [← hf.dom_eq] exact mem_pair_imp_fst_mem_dom ht rw [heq] at this exact single_valued_eq_unique hf.is_func this.right ht /-- #### Theorem 3M If `R` is a symmetric and transitive relation, then `R` is an equivalence relation on `fld R`. -/ theorem theorem_3m {R : Set.Relation α} (hS : R.isSymmetric) (hT : R.isTransitive) : R.isEquivalence (fld R) := by refine ⟨Eq.subset rfl, ?_, hS, hT⟩ intro x hx apply Or.elim hx · intro h have ⟨y, hy⟩ := dom_exists h have := hS hy exact hT hy this · intro h have ⟨t, ht⟩ := ran_exists h have := hS ht exact hT this ht /-- #### Exercise 3.32 (a) Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`. -/ theorem exercise_3_32_a {R : Set.Relation α} : isSymmetric R ↔ inv R ⊆ R := by apply Iff.intro · intro hR show ∀ p, p ∈ inv R → p ∈ R intro (x, y) hp simp only [mem_self_comm_mem_inv] at hp exact hR hp · intro hR unfold isSymmetric intro x y hp rw [← mem_self_comm_mem_inv] at hp exact hR hp /-- #### Exercise 3.32 (b) Show that `R` is transitive **iff** `R ∘ R ⊆ R`. -/ theorem exercise_3_32_b {R : Set.Relation α} : isTransitive R ↔ comp R R ⊆ R := by apply Iff.intro · intro hR show ∀ p, p ∈ comp R R → p ∈ R intro (x, y) hp have ⟨t, ht⟩ := hp exact hR ht.left ht.right · intro hR intro x y z hx hz have : (x, z) ∈ comp R R := ⟨y, hx, hz⟩ exact hR this /-- #### Exercise 3.33 Show that `R` is a symmetric and transitive relation **iff** `R = R⁻¹ ∘ R`. -/ theorem exercise_3_33 {R : Set.Relation α} : isSymmetric R ∧ isTransitive R ↔ R = comp (inv R) R := by have hR : comp (inv R) R = { p | ∃ t, (p.1, t) ∈ R ∧ (p.2, t) ∈ R } := by ext p unfold comp inv simp only [Prod.exists, Set.mem_setOf_eq, Prod.mk.injEq] apply Iff.intro · intro ⟨t, ht, a, b, h⟩ refine ⟨t, ht, ?_⟩ rw [← h.right.right, ← h.right.left] exact h.left · intro ⟨t, ht⟩ exact ⟨t, ht.left, p.snd, t, ht.right, rfl, rfl⟩ apply Iff.intro · intro h rw [Set.Subset.antisymm_iff] apply And.intro · show ∀ p, p ∈ R → p ∈ comp (inv R) R intro (x, y) hp have hy := h.left hp have hx := h.right hp hy rw [hR] exact ⟨x, hx, hy⟩ · show ∀ p, p ∈ comp (inv R) R → p ∈ R intro (x, y) hp rw [hR] at hp have ⟨_, ht⟩ := hp have := h.left ht.right exact h.right ht.left this · intro h have hS : isSymmetric R := by intro x y hp have : inv R = R := by calc inv R _ = inv (comp (inv R) R) := by conv => lhs; rw [h] _ = comp (inv R) (inv (inv R)) := by rw [comp_inv_eq_inv_comp_inv] _ = comp (inv R) R := by rw [inv_inv_eq_self] _ = R := h.symm rwa [← this, mem_self_comm_mem_inv] refine ⟨hS, ?_⟩ intro x y z hx hy have : (z, y) ∈ R := hS hy rw [h, hR] exact ⟨y, hx, this⟩ /-- #### Exercise 3.34 (a) Assume that `𝓐` is a nonempty set, every member of which is a transitive relation. Is the set `⋂ 𝓐` a transitive relation? -/ theorem exercise_3_34_a {𝓐 : Set (Set.Relation α)} (_ : Set.Nonempty 𝓐) (h𝓐 : ∀ A ∈ 𝓐, isTransitive A) : isTransitive (⋂₀ 𝓐) := by intro x y z hx hy simp only [Set.mem_sInter] at * intro A hA have hx' := hx A hA have hy' := hy A hA exact h𝓐 A hA hx' hy' /-- #### Exercise 3.34 (b) Assume that `𝓐` is a nonempty set, every member of which is a transitive relation. Is `⋃ 𝓐` a transitive relation? -/ theorem exercise_3_34_b {𝓐 : Set (Set.Relation ℕ)} (_ : Set.Nonempty 𝓐) (h𝓐 : 𝓐 = {{(1, 2), (2, 3), (1, 3)}, {(2, 1)}}) : (∀ A ∈ 𝓐, isTransitive A) ∧ ¬ isTransitive (⋃₀ 𝓐) := by apply And.intro · intro A hA rw [h𝓐] at hA simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at hA apply Or.elim hA · intro hA₁ rw [hA₁] intro x y z hx hy simp only [Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq] at * casesm* _ ∨ _ all_goals case _ hl hr => first | {rw [hl.right] at hr; simp at hr} | {rw [hl.left] at hr; simp at hr} | {right; right; exact ⟨hl.left, hr.right⟩} · intro hA₁ rw [hA₁] intro x y z hx hy simp only [Set.mem_singleton_iff, Set.mem_insert_iff, Prod.mk.injEq] at * rw [hx.right] at hy simp at hy · intro h have h₁ : (1, 2) ∈ ⋃₀ 𝓐 := by simp only [Set.mem_sUnion] exact ⟨{(1, 2), (2, 3), (1, 3)}, by rw [h𝓐]; simp, by simp⟩ have h₂ : (2, 1) ∈ ⋃₀ 𝓐 := by simp only [Set.mem_sUnion] exact ⟨{(2, 1)}, by rw [h𝓐]; simp, by simp⟩ have h₃ : (1, 1) ∉ ⋃₀ 𝓐 := by simp only [Set.mem_sUnion] rw [h𝓐] intro ⟨t, ht⟩ simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at ht have := ht.right apply Or.elim ht.left <;> · intro ht₁ rw [ht₁] at this simp at this exact absurd (h h₁ h₂) h₃ /-- #### Exercise 3.35 Show that for any `R` and `x`, we have `[x]_R = R⟦{x}⟧`. -/ theorem exercise_3_35 {R : Set.Relation α} {x : α} : neighborhood R x = image R {x} := by calc neighborhood R x _ = { t | (x, t) ∈ R } := rfl _ = { t | ∃ u ∈ ({x} : Set α), (u, t) ∈ R } := by simp _ = image R {x} := rfl /-- #### Exercise 3.36 Assume that `f : A → B` and that `R` is an equivalence relation on `B`. Define `Q` to be the set `{⟨x, y⟩ ∈ A × A | ⟨f(x), f(y)⟩ ∈ R}`. Show that `Q` is an equivalence relation on `A`. -/ theorem exercise_3_36 {f : Set.HRelation α β} {Q : Set.Relation α} {R : Set.Relation β} {A : Set α} {B : Set β} (hf : mapsInto f A B) (hR : isEquivalence R B) (hQ : Q = { p | ∃ fx fy : β, (p.1, fx) ∈ f ∧ (p.2, fy) ∈ f ∧ (fx, fy) ∈ R }) : isEquivalence Q A := by refine ⟨?_, ?_, ?_, ?_⟩ · -- `fld Q ⊆ A` show ∀ x, x ∈ fld Q → x ∈ A intro x hx rw [hQ] at hx unfold fld dom ran at hx simp only [ exists_and_left, Set.mem_union, Set.mem_image, Set.mem_setOf_eq, Prod.exists, exists_and_right, exists_eq_right ] at hx apply Or.elim hx · intro ⟨_, _, hx₁⟩ rw [← hf.dom_eq] exact mem_pair_imp_fst_mem_dom hx₁.left · intro ⟨_, _, _, _, hx₂⟩ rw [← hf.dom_eq] exact mem_pair_imp_fst_mem_dom hx₂.left · -- `isReflexive Q A` intro x hx rw [← hf.dom_eq] at hx have ⟨fx, hfx⟩ := dom_exists hx have := hR.refl fx (hf.ran_ss $ mem_pair_imp_snd_mem_ran hfx) rw [hQ] simp only [exists_and_left, Set.mem_setOf_eq] exact ⟨fx, hfx, fx, hfx, this⟩ · -- `isSymmetric Q` intro x y h rw [hQ] at h simp only [exists_and_left, Set.mem_setOf_eq] at h have ⟨fx, hfx, fy, hfy, h'⟩ := h have := hR.symm h' rw [hQ] simp only [exists_and_left, Set.mem_setOf_eq] exact ⟨fy, hfy, fx, hfx, this⟩ · -- `isTransitive Q` intro x y z hx hy rw [hQ] at hx hy simp only [exists_and_left, Set.mem_setOf_eq] at hx hy have ⟨fx, hfx, fy, hfy, h₁⟩ := hx have ⟨fy₁, hfy₁, fz, hfz, h₂⟩ := hy have hfy' : fy = fy₁ := single_valued_eq_unique hf.is_func hfy hfy₁ rw [hfy'] at h₁ rw [hQ] simp only [exists_and_left, Set.mem_setOf_eq] exact ⟨fx, hfx, fz, hfz, hR.trans h₁ h₂⟩ /-- #### Exercise 3.37 Assume that `P` is a partition of a set `A`. Define the relation `Rₚ` as follows: ``` xRₚy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B). ``` 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) (Rₚ : Set.Relation α) (hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B) : isEquivalence Rₚ A := by have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by ext p have (x, y) := p exact hRₚ x y refine ⟨?_, ?_, ?_, ?_⟩ · -- `fld Rₚ ⊆ A` show ∀ x, x ∈ fld Rₚ → x ∈ A rw [hRₚ_eq] intro x hx unfold fld dom ran at hx simp only [ Set.mem_union, Set.mem_image, Set.mem_setOf_eq, Prod.exists, exists_and_right, exists_eq_right ] at hx apply Or.elim hx · intro ⟨t, B, hB⟩ have := hP.p_subset B hB.left exact this hB.right.left · intro ⟨a, B, hB⟩ have := hP.p_subset B hB.left exact this hB.right.right · -- `isReflexive Rₚ A` intro x hx rw [hRₚ_eq] simp only [Set.mem_setOf_eq, and_self] exact hP.exhaustive x hx · -- `isSymmetric Rₚ` intro x y h rw [hRₚ_eq] at h simp only [Set.mem_setOf_eq] at h have ⟨B, hB⟩ := h rw [hRₚ_eq] simp only [Set.mem_setOf_eq] conv at hB => right; rw [and_comm] exact ⟨B, hB⟩ · -- `isTransitive Rₚ` intro x y z hx hy rw [hRₚ_eq] at hx hy simp only [Set.mem_setOf_eq] at hx hy have ⟨B₁, hB₁⟩ := hx have ⟨B₂, hB₂⟩ := hy have hB : B₁ = B₂ := by have hy₁ : y ∈ B₁ := hB₁.right.right have hy₂ : y ∈ B₂ := hB₂.right.left have hy := hP.disjoint B₁ hB₁.left B₂ hB₂.left rw [contraposition] at hy simp at hy suffices B₁ ∩ B₂ ≠ ∅ from hy this intro h' rw [Set.ext_iff] at h' exact (h' y).mp ⟨hy₁, hy₂⟩ rw [hRₚ_eq] 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 hRₚ_eq : 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, hRₚ_eq] 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, hRₚ_eq] 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 [hRₚ_eq] simp only [Set.mem_setOf_eq] _ = neighborhood Rₚ x := rfl /-- #### Exercise 3.39 Assume that we start with an equivalence relation `R` on `A` and define `P` to be the partition `A / R`. Show that `Rₚ`, as defined in Exercise 37, is just `R`. -/ theorem exercise_3_39 {P : Set (Set α)} {R Rₚ : Set.Relation α} {A : Set α} (hR : isEquivalence R A) (hP : P = modEquiv hR) (hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B) : Rₚ = R := by have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by ext p have (x, y) := p exact hRₚ x y ext p have (x, y) := p apply Iff.intro · intro hp rw [hRₚ_eq] at hp have ⟨B, hB, hx, hy⟩ := hp rw [hP] at hB have ⟨z, hz⟩ := hB rw [← hz.right] at hx hy exact neighborhood_mem_imp_relate hR hx hy · intro hp have hxA : x ∈ A := hR.b_on (Or.inl (mem_pair_imp_fst_mem_dom hp)) have hyA : y ∈ A := hR.b_on (Or.inr (mem_pair_imp_snd_mem_ran hp)) rw [hRₚ_eq] have hx : x ∈ neighborhood R x := neighborhood_self_mem hR hxA have hy : y ∈ neighborhood R x := by rw [← neighborhood_eq_iff_mem_relate hR hxA hyA] at hp rw [hp] exact neighborhood_self_mem hR hyA refine ⟨neighborhood R x, ?_, ⟨hx, hy⟩⟩ rw [hP] exact ⟨x, hxA, rfl⟩ /-- #### Exercise 3.41 (a) Let `ℝ` be the set of real numbers and define the realtion `Q` on `ℝ × ℝ` by `⟨u, v⟩ Q ⟨x, y⟩` **iff** `u + y = x + v`. Show that `Q` is an equivalence relation on `ℝ × ℝ`. -/ theorem exercise_3_41_a {Q : Set.Relation (ℝ × ℝ)} (hQ : ∀ u v x y, ((u, v), (x, y)) ∈ Q ↔ u + y = x + v) : isEquivalence Q (Set.univ : Set (ℝ × ℝ)) := by have hQ_eq : Q = {p | p.1.1 + p.2.2 = p.2.1 + p.1.2} := by ext p apply Iff.intro <;> · intro hp rwa [hQ] at * refine ⟨?_, ?_, ?_, ?_⟩ · -- `fld Q ⊆ Set.univ` show ∀ p, p ∈ fld Q → p ∈ Set.univ intro _ _ simp only [Set.mem_univ] · -- `isReflexive Q Set.univ` intro (x, y) _ rw [hQ_eq] simp · -- `isSymmetric Q` intro (u, v) (x, y) hp rw [hQ_eq] at * exact Eq.symm hp · -- `isTransitive Q` unfold isTransitive intro (u, v) (x, y) (a, b) h₁ h₂ rw [hQ_eq] at * simp at h₁ h₂ simp have h₁' : u - v = x - y := by have := sub_eq_of_eq_add h₁ rw [add_sub_right_comm] at this exact eq_sub_of_add_eq this have h₂' : x - y = a - b := by have := sub_eq_of_eq_add h₂ rw [add_sub_right_comm] at this exact eq_sub_of_add_eq this rw [h₂'] at h₁' have := eq_add_of_sub_eq' h₁' rw [← add_sub_assoc] at this have := add_eq_of_eq_sub this conv => right; rw [add_comm] exact this end Relation /-- #### Theorem 3R Let `R` be a linear ordering on `A`. (i) There is no `x` for which `xRx`. (ii) For distinct `x` and `y` in `A`, either `xRy` or `yRx`. -/ theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R) : (∀ x : α, ¬ R x x) ∧ (∀ x y : α, x ≠ y → R x y ∨ R y x) := by apply And.intro · exact hR.irrefl · intro x y h apply Or.elim (hR.trichotomous x y) · intro h₁ left exact h₁ · intro h₁ apply Or.elim h₁ · intro h₂ exact absurd h₂ h · intro h₂ right exact h₂ /-- #### Exercise 3.43 Assume that `R` is a linear ordering on a set `A`. Show that `R⁻¹` is also a linear ordering on `A`. -/ theorem exercise_3_43 {R : Rel α α} (hR : IsStrictTotalOrder α R) : IsStrictTotalOrder α R.inv := by refine { trichotomous := ?_, irrefl := ?_, trans := ?_ } · intro a b unfold Rel.inv flip apply Or.elim (hR.trichotomous a b) · intro h; right; right; exact h · intro h apply Or.elim h · intro h; right; left; exact h · intro h; left; exact h · intro x h unfold Rel.inv flip at h exact absurd h (hR.irrefl x) · intro a b c hab hac unfold Rel.inv flip at * exact hR.trans c b a hac hab /-! #### Exercise 3.44 Assume that `<` is a linear ordering on a set `A`. Assume that `f : A → A` and that `f` has the property that whenever `x < y`, then `f(x) < f(y)`. Show that `f` is one-to-one and that whenever `f(x) < f(y)`, then `x < y`. -/ theorem exercise_3_44_i {R : Rel α α} (hR : IsStrictTotalOrder α R) (f : α → α) (hf : ∀ x y, R x y → R (f x) (f y)) : Function.Injective f := by unfold Function.Injective intro x₁ x₂ hx apply Or.elim (hR.trichotomous x₁ x₂) · -- `x₁ < x₂` intro hx₁ have nh := hf x₁ x₂ hx₁ rw [hx] at nh exact absurd nh (hR.irrefl (f x₂)) · intro hx₁ apply Or.elim hx₁ · simp -- `x₁ = x₂` · -- `x₁ > x₂` intro hx₂ have nh := hf x₂ x₁ hx₂ rw [← hx] at nh exact absurd nh (hR.irrefl (f x₁)) theorem exercise_3_44_ii {R : Rel α α} (hR : IsStrictTotalOrder α R) (f : α → α) (hf : ∀ x y, R x y → R (f x) (f y)) : R (f x) (f y) → R x y := by intro h apply Or.elim (hR.trichotomous x y) · simp -- `x < y` · intro h₁ apply Or.elim h₁ · -- `x = y` intro h₂ rw [h₂] at h exact absurd h (hR.irrefl (f y)) · -- `x > y` intro h₂ have := hR.trans (f x) (f y) (f x) h (hf y x h₂) exact absurd this (hR.irrefl (f x)) /-- #### Exercise 3.45 Assume that `<_A` and `<_B` are linear orderings on `A` and `B`, respectively. Define the binary relation `<_L` on the Cartesian product `A × B` by: ``` ⟨a₁, b₁⟩ <_L ⟨a₂, b₂⟩ iff either a₁ <_A a₂ or (a₁ = a₂ ∧ b₁ <_B b₂). ``` Show that `<_L` is a linear ordering on `A × B`. (The relation `<_L` is called *lexicographic* ordering, being the ordering used in making dictionaries.) -/ theorem exercise_3_45 {A : Rel α α} {B : Rel β β} {R : Rel (α × β) (α × β)} (hA : IsStrictTotalOrder α A) (hB : IsStrictTotalOrder β B) (hR : ∀ a₁ b₁ a₂ b₂, R (a₁, b₁) (a₂, b₂) ↔ A a₁ a₂ ∨ (a₁ = a₂ ∧ B b₁ b₂)) : IsStrictTotalOrder (α × β) R := by refine { trichotomous := ?_, irrefl := ?_, trans := ?_ } · intro (a₁, b₁) (a₂, b₂) apply Or.elim (hA.trichotomous a₁ a₂) · -- `a₁ <_A a₂` intro ha left exact (hR a₁ b₁ a₂ b₂).mpr (Or.inl ha) · intro ha apply Or.elim ha · -- `a₁ = a₂` intro ha₁ apply Or.elim (hB.trichotomous b₁ b₂) · -- `b₁ <_B b₂` intro hb left exact (hR a₁ b₁ a₂ b₂).mpr (Or.inr ⟨ha₁, hb⟩) · intro hb apply Or.elim hb · -- `b₁ = b₂` intro hb₁ right; left rw [ha₁, hb₁] · -- `b₂ <_B b₁` intro hb₁ right; right exact (hR a₂ b₂ a₁ b₁).mpr (Or.inr ⟨ha₁.symm, hb₁⟩) · -- `a₂ <_A a₁` intro ha₁ right; right exact (hR a₂ b₂ a₁ b₁).mpr (Or.inl ha₁) · intro (a, b) h have := (hR a b a b).mp h apply Or.elim this · intro ha₁ exact absurd ha₁ (hA.irrefl a) · intro ⟨_, hb₁⟩ exact absurd hb₁ (hB.irrefl b) · intro (a₁, b₁) (a₂, b₂) (a₃, b₃) h₁ h₂ have h₁' := (hR a₁ b₁ a₂ b₂).mp h₁ have h₂' := (hR a₂ b₂ a₃ b₃).mp h₂ apply Or.elim h₁' · -- `a₁ <_A a₂` intro ha₁ apply Or.elim h₂' · -- `a₂ <_A a₃` intro ha₂ have := hA.trans a₁ a₂ a₃ ha₁ ha₂ exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl this) · -- `a₂ = a₃ ∧ b₂ <_B b₃` intro ha₂ rw [ha₂.left] at ha₁ exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl ha₁) · -- `a₁ = a₂ ∧ b₁ <_B b₂` intro ha₁ apply Or.elim h₂' · -- `a₂ <_A a₃` intro ha₂ rw [← ha₁.left] at ha₂ exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl ha₂) · -- `a₂ = a₃ ∧ b₂ <_B b₃` intro ⟨ha₂, hb₂⟩ rw [← ha₁.left] at ha₂ have := hB.trans b₁ b₂ b₃ ha₁.right hb₂ exact (hR a₁ b₁ a₃ b₃).mpr (Or.inr ⟨ha₂, this⟩) end Enderton.Set.Chapter_3