import Bookshelf.Enderton.Set.Chapter_2 import Bookshelf.Enderton.Set.OrderedPair import Bookshelf.Enderton.Set.Relation /-! # Enderton.Set.Chapter_3 Relations and Functions -/ namespace Enderton.Set.Chapter_3 /-- #### Theorem 3B If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`. -/ theorem theorem_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)⟩ /-- #### Exercise 3.6 Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`. -/ theorem exercise_3_6 {A : Set.Relation α} : A ⊆ Set.prod (A.dom) (A.ran) := 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 (R.dom) (R.ran) := exercise_3_6 have : (x, y) ∈ Set.prod (R.dom) (R.ran) := 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 section open Set.Relation /-- #### Exercise 3.8 (i) Show that for any set `𝓐`: ``` dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 } ``` -/ theorem exercise_3_8_i {A : Set (Set.Relation α)} : 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.Relation α)} : 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.Relation α)} : 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.Relation α)} : 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.Relation α} (hF : F.isOneToOne) (hx : x ∈ dom F) : ∃! y, (x, y) ∈ F ∧ (y, x) ∈ F.inv := 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.Relation α} (hF : F.isOneToOne) (hy : y ∈ F.ran) : ∃! x, (x, y) ∈ F ∧ (y, x) ∈ F.inv := 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 G : Set.Relation α} (_ : F.isSingleValued) (hG : G.isSingleValued) : dom (F.comp 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 (F.comp 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 (F.comp 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.Relation α} {A B : Set α} (hF : F.mapsInto A B) (hA : Set.Nonempty A) : (∃ G : Set.Relation α, G.mapsInto B A ∧ (∀ p ∈ G.comp F, p.1 = p.2)) ↔ F.isOneToOne := by 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.Relation α} {A B : Set α} (hF : F.mapsInto A B) (hA : Set.Nonempty A) : (∃ H : Set.Relation α, H.mapsInto B A ∧ (∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by sorry end end Enderton.Set.Chapter_3