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 (α ⊕ α)} (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 : 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 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 (α ⊕ α)))} (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