Enderton. Fix exercise names.
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@ -13,13 +13,13 @@ Axioms and Operations
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namespace Enderton.Set.Chapter_2
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/-- ### Exercise 3.1
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/-- ### Exercise 2.1
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Assume that `A` is the set of integers divisible by `4`. Similarly assume that
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`B` and `C` are the sets of integers divisible by `9` and `10`, respectively.
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What is in `A ∩ B ∩ C`?
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-/
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theorem exercise_3_1 {A B C : Set ℤ}
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theorem exercise_2_1 {A B C : Set ℤ}
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(hA : A = { x | Dvd.dvd 4 x })
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(hB : B = { x | Dvd.dvd 9 x })
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(hC : C = { x | Dvd.dvd 10 x })
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@ -35,11 +35,11 @@ theorem exercise_3_1 {A B C : Set ℤ}
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· rw [hC] at hc
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exact Set.mem_setOf.mp hc
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/-- ### Exercise 3.2
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/-- ### Exercise 2.2
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Give an example of sets `A` and `B` for which `⋃ A = ⋃ B` but `A ≠ B`.
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-/
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theorem exercise_3_2 {A B : Set (Set ℕ)}
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theorem exercise_2_2 {A B : Set (Set ℕ)}
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(hA : A = {{1}, {2}}) (hB : B = {{1, 2}})
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: Set.sUnion A = Set.sUnion B ∧ A ≠ B := by
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apply And.intro
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@ -74,12 +74,12 @@ theorem exercise_3_2 {A B : Set (Set ℕ)}
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have h₂ := h₁ 2
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simp at h₂
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/-- ### Exercise 3.3
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/-- ### Exercise 2.3
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Show that every member of a set `A` is a subset of `U A`. (This was stated as an
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example in this section.)
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-/
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theorem exercise_3_3 {A : Set (Set α)}
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theorem exercise_2_3 {A : Set (Set α)}
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: ∀ x ∈ A, x ⊆ ⋃₀ A := by
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intro x hx
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show ∀ y ∈ x, y ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }
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@ -87,11 +87,11 @@ theorem exercise_3_3 {A : Set (Set α)}
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rw [Set.mem_setOf_eq]
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exact ⟨x, ⟨hx, hy⟩⟩
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/-- ### Exercise 3.4
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/-- ### Exercise 2.4
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Show that if `A ⊆ B`, then `⋃ A ⊆ ⋃ B`.
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-/
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theorem exercise_3_4 {A B : Set (Set α)} (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
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theorem exercise_2_4 {A B : Set (Set α)} (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B := by
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show ∀ x ∈ { a | ∃ t, t ∈ A ∧ a ∈ t }, x ∈ { a | ∃ t, t ∈ B ∧ a ∈ t }
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intro x hx
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rw [Set.mem_setOf_eq] at hx
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@ -99,11 +99,11 @@ theorem exercise_3_4 {A B : Set (Set α)} (h : A ⊆ B) : ⋃₀ A ⊆ ⋃₀ B
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rw [Set.mem_setOf_eq]
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exact ⟨t, ⟨h ht, hxt⟩⟩
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/-- ### Exercise 3.5
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/-- ### Exercise 2.5
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Assume that every member of `𝓐` is a subset of `B`. Show that `⋃ 𝓐 ⊆ B`.
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-/
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theorem exercise_3_5 {𝓐 : Set (Set α)} (h : ∀ x ∈ 𝓐, x ⊆ B)
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theorem exercise_2_5 {𝓐 : Set (Set α)} (h : ∀ x ∈ 𝓐, x ⊆ B)
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: ⋃₀ 𝓐 ⊆ B := by
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show ∀ y ∈ { a | ∃ t, t ∈ 𝓐 ∧ a ∈ t }, y ∈ B
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intro y hy
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@ -111,11 +111,11 @@ theorem exercise_3_5 {𝓐 : Set (Set α)} (h : ∀ x ∈ 𝓐, x ⊆ B)
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have ⟨t, ⟨ht𝓐, hyt⟩⟩ := hy
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exact (h t ht𝓐) hyt
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/-- ### Exercise 3.6a
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/-- ### Exercise 2.6a
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Show that for any set `A`, `⋃ 𝓟 A = A`.
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-/
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theorem exercise_3_6a : ⋃₀ (𝒫 A) = A := by
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theorem exercise_2_6a : ⋃₀ (𝒫 A) = A := by
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show { a | ∃ t, t ∈ { t | t ⊆ A } ∧ a ∈ t } = A
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ext x
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apply Iff.intro
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@ -128,30 +128,30 @@ theorem exercise_3_6a : ⋃₀ (𝒫 A) = A := by
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rw [Set.mem_setOf_eq]
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exact ⟨A, ⟨by rw [Set.mem_setOf_eq], hx⟩⟩
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/-- ### Exercise 3.6b
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/-- ### Exercise 2.6b
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Show that `A ⊆ 𝓟 ⋃ A`. Under what conditions does equality hold?
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-/
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theorem exercise_3_6b
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theorem exercise_2_6b
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: A ⊆ 𝒫 (⋃₀ A)
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∧ (A = 𝒫 (⋃₀ A) ↔ ∃ B, A = 𝒫 B) := by
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apply And.intro
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· show ∀ x ∈ A, x ∈ { t | t ⊆ ⋃₀ A }
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intro x hx
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rw [Set.mem_setOf]
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exact exercise_3_3 x hx
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exact exercise_2_3 x hx
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· apply Iff.intro
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· intro hA
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exact ⟨⋃₀ A, hA⟩
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· intro ⟨B, hB⟩
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conv => rhs; rw [hB, exercise_3_6a]
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conv => rhs; rw [hB, exercise_2_6a]
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exact hB
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/-- ### Exercise 3.7a
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/-- ### Exercise 2.7a
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Show that for any sets `A` and `B`, `𝓟 A ∩ 𝓟 B = 𝓟 (A ∩ B)`.
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-/
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theorem exercise_3_7A
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theorem exercise_2_7A
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: 𝒫 A ∩ 𝒫 B = 𝒫 (A ∩ B) := by
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suffices 𝒫 A ∩ 𝒫 B ⊆ 𝒫 (A ∩ B) ∧ 𝒫 (A ∩ B) ⊆ 𝒫 A ∩ 𝒫 B from
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subset_antisymm this.left this.right
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@ -167,11 +167,11 @@ theorem exercise_3_7A
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-- theorem false_of_false_iff_true : (false ↔ true) → false := by simp
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/-- ### Exercise 3.7b (i)
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/-- ### Exercise 2.7b (i)
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Show that `𝓟 A ∪ 𝓟 B ⊆ 𝓟 (A ∪ B)`.
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-/
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theorem exercise_3_7b_i
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theorem exercise_2_7b_i
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: 𝒫 A ∪ 𝒫 B ⊆ 𝒫 (A ∪ B) := by
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unfold Set.powerset
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intro x hx
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@ -184,11 +184,11 @@ theorem exercise_3_7b_i
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rw [Set.mem_setOf_eq]
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exact Set.subset_union_of_subset_right hB A
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/-- ### Exercise 3.7b (ii)
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/-- ### Exercise 2.7b (ii)
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Under what conditions does `𝓟 A ∪ 𝓟 B = 𝓟 (A ∪ B)`.?
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-/
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theorem exercise_3_7b_ii
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theorem exercise_2_7b_ii
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: 𝒫 A ∪ 𝒫 B = 𝒫 (A ∪ B) ↔ A ⊆ B ∨ B ⊆ A := by
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unfold Set.powerset
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apply Iff.intro
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@ -236,11 +236,11 @@ theorem exercise_3_7b_ii
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refine Or.inl (Set.Subset.trans hx ?_)
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exact subset_of_eq (Set.right_subset_union_eq_self hB)
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/-- ### Exercise 3.9
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/-- ### Exercise 2.9
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Give an example of sets `a` and `B` for which `a ∈ B` but `𝓟 a ∉ 𝓟 B`.
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-/
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theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
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theorem exercise_2_9 (ha : a = {1}) (hB : B = {{1}})
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: a ∈ B ∧ 𝒫 a ∉ 𝒫 B := by
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apply And.intro
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· rw [ha, hB]
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@ -264,24 +264,24 @@ theorem exercise_3_9 (ha : a = {1}) (hB : B = {{1}})
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have := h 1
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simp at this
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/-- ### Exercise 3.10
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/-- ### Exercise 2.10
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Show that if `a ∈ B`, then `𝓟 a ∈ 𝓟 𝓟 ⋃ B`.
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-/
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theorem exercise_3_10 {B : Set (Set α)} (ha : a ∈ B)
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theorem exercise_2_10 {B : Set (Set α)} (ha : a ∈ B)
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: 𝒫 a ∈ 𝒫 (𝒫 (⋃₀ B)) := by
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have h₁ := exercise_3_3 a ha
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have h₁ := exercise_2_3 a ha
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have h₂ := Chapter_1.exercise_1_3 h₁
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generalize hb : 𝒫 (⋃₀ B) = b
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conv => rhs; unfold Set.powerset
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rw [← hb, Set.mem_setOf_eq]
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exact h₂
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/-- ### Exercise 4.11 (i)
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/-- ### Exercise 2.11 (i)
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Show that for any sets `A` and `B`, `A = (A ∩ B) ∪ (A - B)`.
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-/
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theorem exercise_4_11_i {A B : Set α}
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theorem exercise_2_11_i {A B : Set α}
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: A = (A ∩ B) ∪ (A \ B) := by
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show A = fun a => A a ∧ B a ∨ A a ∧ ¬B a
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suffices ∀ x, (A x ∧ (B x ∨ ¬B x)) = A x by
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@ -294,11 +294,11 @@ theorem exercise_4_11_i {A B : Set α}
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· intro hx
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exact ⟨hx, em (B x)⟩
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/-- ### Exercise 4.11 (ii)
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/-- ### Exercise 2.11 (ii)
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Show that for any sets `A` and `B`, `A ∪ (B - A) = A ∪ B`.
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-/
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theorem exercise_4_11_ii {A B : Set α}
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theorem exercise_2_11_ii {A B : Set α}
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: A ∪ (B \ A) = A ∪ B := by
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show (fun a => A a ∨ B a ∧ ¬A a) = fun a => A a ∨ B a
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suffices ∀ x, ((A x ∨ B x) ∧ (A x ∨ ¬A x)) = (A x ∨ B x) by
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@ -388,12 +388,12 @@ lemma left_diff_eq_singleton_one : (A \ B) \ C = {1} := by
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| inl y => rw [hx] at y; simp at y
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| inr y => rw [hx] at y; simp at y
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/-- ### Exercise 4.14
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/-- ### Exercise 2.14
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Show by example that for some sets `A`, `B`, and `C`, the set `A - (B - C)` is
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different from `(A - B) - C`.
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-/
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theorem exercise_4_14 : A \ (B \ C) ≠ (A \ B) \ C := by
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theorem exercise_2_14 : A \ (B \ C) ≠ (A \ B) \ C := by
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rw [
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@right_diff_eq_insert_one_three A B C hA hB hC,
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@left_diff_eq_singleton_one A B C hA hB hC
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@ -405,19 +405,19 @@ theorem exercise_4_14 : A \ (B \ C) ≠ (A \ B) \ C := by
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end
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/-- ### Exercise 4.16
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/-- ### Exercise 2.16
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Simplify:
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`[(A ∪ B ∪ C) ∩ (A ∪ B)] - [(A ∪ (B - C)) ∩ A]`
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-/
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theorem exercise_4_16 {A B C : Set α}
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theorem exercise_2_16 {A B C : Set α}
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: ((A ∪ B ∪ C) ∩ (A ∪ B)) \ ((A ∪ (B \ C)) ∩ A) = B \ A := by
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calc ((A ∪ B ∪ C) ∩ (A ∪ B)) \ ((A ∪ (B \ C)) ∩ A)
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_ = (A ∪ B) \ ((A ∪ (B \ C)) ∩ A) := by rw [Set.union_inter_cancel_left]
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_ = (A ∪ B) \ A := by rw [Set.union_inter_cancel_left]
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_ = B \ A := by rw [Set.union_diff_left]
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/-! ### Exercise 4.17
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/-! ### Exercise 2.17
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Show that the following four conditions are equivalent.
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@ -427,7 +427,7 @@ Show that the following four conditions are equivalent.
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(d) `A ∩ B = A`
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-/
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theorem exercise_4_17_i {A B : Set α} (h : A ⊆ B)
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theorem exercise_2_17_i {A B : Set α} (h : A ⊆ B)
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: A \ B = ∅ := by
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ext x
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apply Iff.intro
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@ -436,7 +436,7 @@ theorem exercise_4_17_i {A B : Set α} (h : A ⊆ B)
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· intro hx
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exact False.elim hx
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theorem exercise_4_17_ii {A B : Set α} (h : A \ B = ∅)
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theorem exercise_2_17_ii {A B : Set α} (h : A \ B = ∅)
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: A ∪ B = B := by
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suffices A ⊆ B from Set.left_subset_union_eq_self this
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show ∀ t, t ∈ A → t ∈ B
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@ -445,19 +445,19 @@ theorem exercise_4_17_ii {A B : Set α} (h : A \ B = ∅)
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by_contra nt
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exact (h t).mp ⟨ht, nt⟩
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theorem exercise_4_17_iii {A B : Set α} (h : A ∪ B = B)
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theorem exercise_2_17_iii {A B : Set α} (h : A ∪ B = B)
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: A ∩ B = A := by
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suffices A ⊆ B from Set.inter_eq_left_iff_subset.mpr this
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exact Set.union_eq_right_iff_subset.mp h
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theorem exercise_4_17_iv {A B : Set α} (h : A ∩ B = A)
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theorem exercise_2_17_iv {A B : Set α} (h : A ∩ B = A)
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: A ⊆ B := Set.inter_eq_left_iff_subset.mp h
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/-- ### Exercise 4.19
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/-- ### Exercise 2.19
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Is `𝒫 (A - B)` always equal to `𝒫 A - 𝒫 B`? Is it ever equal to `𝒫 A - 𝒫 B`?
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-/
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theorem exercise_4_19 {A B : Set α}
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theorem exercise_2_19 {A B : Set α}
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: 𝒫 (A \ B) ≠ (𝒫 A) \ (𝒫 B) := by
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intro h
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have he : ∅ ∈ 𝒫 (A \ B) := by simp
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@ -466,12 +466,12 @@ theorem exercise_4_19 {A B : Set α}
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have := h ∅
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exact absurd (this.mp he) ne
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/-- ### Exercise 4.20
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/-- ### Exercise 2.20
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Let `A`, `B`, and `C` be sets such that `A ∪ B = A ∪ C` and `A ∩ B = A ∩ C`.
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Show that `B = C`.
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-/
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theorem exercise_4_20 {A B C : Set α}
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theorem exercise_2_20 {A B C : Set α}
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(hu : A ∪ B = A ∪ C) (hi : A ∩ B = A ∩ C) : B = C := by
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ext x
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apply Iff.intro
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@ -492,11 +492,11 @@ theorem exercise_4_20 {A B C : Set α}
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rw [← hu] at this
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exact Or.elim this (absurd · hA) (by simp)
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/-- ### Exercise 4.21
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/-- ### Exercise 2.21
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Show that `⋃ (A ∪ B) = (⋃ A) ∪ (⋃ B)`.
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-/
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theorem exercise_4_21 {A B : Set (Set α)}
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theorem exercise_2_21 {A B : Set (Set α)}
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: ⋃₀ (A ∪ B) = (⋃₀ A) ∪ (⋃₀ B) := by
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ext x
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apply Iff.intro
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@ -516,11 +516,11 @@ theorem exercise_4_21 {A B : Set (Set α)}
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have ⟨t, ht⟩ : ∃ t, t ∈ B ∧ x ∈ t := hB
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exact ⟨t, ⟨Set.mem_union_right A ht.left, ht.right⟩⟩
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/-- ### Exercise 4.22
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/-- ### Exercise 2.22
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Show that if `A` and `B` are nonempty sets, then `⋂ (A ∪ B) = ⋂ A ∩ ⋂ B`.
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-/
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theorem exercise_4_22 {A B : Set (Set α)}
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theorem exercise_2_22 {A B : Set (Set α)}
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: ⋂₀ (A ∪ B) = ⋂₀ A ∩ ⋂₀ B := by
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ext x
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apply Iff.intro
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@ -545,11 +545,11 @@ theorem exercise_4_22 {A B : Set (Set α)}
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· intro hB
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exact (this t).right hB
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/-- ### Exercise 4.24a
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/-- ### Exercise 2.24a
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Show that is `𝓐` is nonempty, then `𝒫 (⋂ 𝓐) = ⋂ { 𝒫 X | X ∈ 𝓐 }`.
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-/
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theorem exercise_4_24a {𝓐 : Set (Set α)}
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theorem exercise_2_24a {𝓐 : Set (Set α)}
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: 𝒫 (⋂₀ 𝓐) = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := by
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calc 𝒫 (⋂₀ 𝓐)
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_ = { x | x ⊆ ⋂₀ 𝓐 } := rfl
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@ -564,7 +564,7 @@ theorem exercise_4_24a {𝓐 : Set (Set α)}
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_ = { x | ∀ t ∈ { 𝒫 X | X ∈ 𝓐 }, x ∈ t} := by simp
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_ = ⋂₀ { 𝒫 X | X ∈ 𝓐 } := rfl
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/-- ### Exercise 4.24b
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/-- ### Exercise 2.24b
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Show that
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```
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@ -572,12 +572,12 @@ Show that
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```
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Under what conditions does equality hold?
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-/
|
||||
theorem exercise_4_24b {𝓐 : Set (Set α)}
|
||||
theorem exercise_2_24b {𝓐 : Set (Set α)}
|
||||
: (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐)
|
||||
∧ ((⋃₀ { 𝒫 X | X ∈ 𝓐 } = 𝒫 ⋃₀ 𝓐) ↔ (⋃₀ 𝓐 ∈ 𝓐)) := by
|
||||
have hS : (⋃₀ { 𝒫 X | X ∈ 𝓐 } ⊆ 𝒫 ⋃₀ 𝓐) := by
|
||||
simp
|
||||
exact exercise_3_3
|
||||
exact exercise_2_3
|
||||
refine ⟨hS, ?_⟩
|
||||
apply Iff.intro
|
||||
· intro rS
|
||||
|
|
@ -593,7 +593,7 @@ theorem exercise_4_24b {𝓐 : Set (Set α)}
|
|||
have : ⋃₀ 𝓐 = x := by
|
||||
rw [← hx.right] at ht
|
||||
have hl : ⋃₀ 𝓐 ⊆ x := ht
|
||||
have hr : x ⊆ ⋃₀ 𝓐 := exercise_3_3 x hx.left
|
||||
have hr : x ⊆ ⋃₀ 𝓐 := exercise_2_3 x hx.left
|
||||
exact Set.Subset.antisymm hl hr
|
||||
rw [← this] at hx
|
||||
exact hx.left
|
||||
|
|
@ -606,12 +606,12 @@ theorem exercise_4_24b {𝓐 : Set (Set α)}
|
|||
simp only [Set.mem_setOf_eq, exists_exists_and_eq_and, Set.mem_powerset_iff]
|
||||
exact ⟨⋃₀ 𝓐, ⟨hA, hx⟩⟩
|
||||
|
||||
/-- ### Exercise 4.25
|
||||
/-- ### Exercise 2.25
|
||||
|
||||
Is `A ∪ (⋃ 𝓑)` always the same as `⋃ { A ∪ X | X ∈ 𝓑 }`? If not, then under
|
||||
what conditions does equality hold?
|
||||
-/
|
||||
theorem exercise_4_25 {A : Set α} (𝓑 : Set (Set α))
|
||||
theorem exercise_2_25 {A : Set α} (𝓑 : Set (Set α))
|
||||
: (A ∪ (⋃₀ 𝓑) = ⋃₀ { A ∪ X | X ∈ 𝓑 }) ↔ (A = ∅ ∨ Set.Nonempty 𝓑) := by
|
||||
apply Iff.intro
|
||||
· intro h
|
||||
|
|
|
|||
|
|
@ -19,7 +19,7 @@ theorem theorem_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
|
|||
have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
|
||||
exact Set.mem_mem_imp_pair_subset hxs hxys
|
||||
|
||||
/-- ### Exercise 5.1
|
||||
/-- ### Exercise 3.1
|
||||
|
||||
Suppose that we attempted to generalize the Kuratowski definitions of ordered
|
||||
pairs to ordered triples by defining
|
||||
|
|
@ -31,7 +31,7 @@ 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}
|
||||
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}}
|
||||
|
|
@ -42,11 +42,11 @@ theorem exercise_5_1 {x y z u v w : ℕ}
|
|||
· rw [hy, hv]
|
||||
simp only
|
||||
|
||||
/-- ### Exercise 5.2a
|
||||
/-- ### Exercise 3.2a
|
||||
|
||||
Show that `A × (B ∪ C) = (A × B) ∪ (A × C)`.
|
||||
-/
|
||||
theorem exercise_5_2a {A : Set α} {B C : Set β}
|
||||
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
|
||||
|
|
@ -58,11 +58,11 @@ theorem exercise_5_2a {A : Set α} {B C : Set β}
|
|||
_ = { p | p ∈ Set.prod A B ∨ (p ∈ Set.prod A C) } := rfl
|
||||
_ = (Set.prod A B) ∪ (Set.prod A C) := rfl
|
||||
|
||||
/-- ### Exercise 5.2b
|
||||
/-- ### Exercise 3.2b
|
||||
|
||||
Show that if `A × B = A × C` and `A ≠ ∅`, then `B = C`.
|
||||
-/
|
||||
theorem exercise_5_2b {A : Set α} {B C : Set β}
|
||||
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
|
||||
|
|
@ -87,11 +87,11 @@ theorem exercise_5_2b {A : Set α} {B C : Set β}
|
|||
have ⟨c, hc⟩ := Set.nonempty_iff_ne_empty.mpr (Ne.symm nC)
|
||||
exact (h (a, c)).mpr ⟨ha, hc⟩
|
||||
|
||||
/-- ### Exercise 5.3
|
||||
/-- ### Exercise 3.3
|
||||
|
||||
Show that `A × ⋃ 𝓑 = ⋃ {A × X | X ∈ 𝓑}`.
|
||||
-/
|
||||
theorem exercise_5_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
|
||||
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
|
||||
|
|
@ -115,7 +115,7 @@ theorem exercise_5_3 {A : Set (Set α)} {𝓑 : Set (Set β)}
|
|||
· intro ⟨b, h₁, h₂, h₃⟩
|
||||
exact ⟨b, h₁, h₂, h₃⟩
|
||||
|
||||
/-- ### Exercise 5.5a
|
||||
/-- ### Exercise 3.5a
|
||||
|
||||
Assume that `A` and `B` are given sets, and show that there exists a set `C`
|
||||
such that for any `y`,
|
||||
|
|
@ -124,7 +124,7 @@ 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}
|
||||
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))}
|
||||
|
|
@ -183,11 +183,11 @@ theorem exercise_5_5a {A : Set α} {B : Set β}
|
|||
rw [hab.right]
|
||||
exact ⟨hab.left, hb⟩
|
||||
|
||||
/-- ### Exercise 5.5b
|
||||
/-- ### Exercise 3.5b
|
||||
|
||||
With `A`, `B`, and `C` as above, show that `A × B = ∪ C`.
|
||||
-/
|
||||
theorem exercise_5_5b {A : Set α} (B : Set β)
|
||||
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
|
||||
|
|
@ -222,17 +222,17 @@ 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_3_3 (OrderedPair x y) h
|
||||
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_3_3 {x, y} hq
|
||||
have : {x, y} ⊆ ⋃₀ ⋃₀ A := Chapter_2.exercise_2_3 {x, y} hq
|
||||
exact ⟨this (by simp), this (by simp)⟩
|
||||
|
||||
/-- ### Exercise 6.6
|
||||
/-- ### Exercise 3.6
|
||||
|
||||
Show that a set `A` is a relation **iff** `A ⊆ dom A × ran A`.
|
||||
-/
|
||||
theorem exercise_6_6 {A : Set.Relation α}
|
||||
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
|
||||
|
|
@ -246,11 +246,11 @@ theorem exercise_6_6 {A : Set.Relation α}
|
|||
]
|
||||
exact ⟨⟨b, ht⟩, ⟨a, ht⟩⟩
|
||||
|
||||
/-- ### Exercise 6.7
|
||||
/-- ### Exercise 3.7
|
||||
|
||||
Show that if `R` is a relation, then `fld R = ⋃ ⋃ R`.
|
||||
-/
|
||||
theorem exercise_6_7 {R : Set.Relation α}
|
||||
theorem exercise_3_7 {R : Set.Relation α}
|
||||
: R.fld = ⋃₀ ⋃₀ R.toOrderedPairs := by
|
||||
let img := R.toOrderedPairs
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
|
|
@ -269,8 +269,8 @@ theorem exercise_6_7 {R : Set.Relation α}
|
|||
simp only [Prod.exists, Set.mem_setOf_eq]
|
||||
exact ⟨x, ⟨y, ⟨hp, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {x} ∈ ⋃₀ img := Chapter_2.exercise_3_3 {{x}, {x, y}} hm (by simp)
|
||||
exact (Chapter_2.exercise_3_3 {x} this) (show x ∈ {x} by rfl)
|
||||
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
|
||||
|
|
@ -279,9 +279,9 @@ theorem exercise_6_7 {R : Set.Relation α}
|
|||
simp only [Set.mem_image, Prod.exists]
|
||||
exact ⟨t, ⟨x, ⟨ht, rfl⟩⟩⟩
|
||||
unfold OrderedPair at hm
|
||||
have : {t, x} ∈ ⋃₀ img := Chapter_2.exercise_3_3 {{t}, {t, x}} 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_3_3 {t, x} this (show x ∈ {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
|
||||
|
|
@ -300,7 +300,7 @@ theorem exercise_6_7 {R : Set.Relation α}
|
|||
-- `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_6_6
|
||||
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
|
||||
|
|
@ -329,14 +329,14 @@ section
|
|||
|
||||
open Set.Relation
|
||||
|
||||
/-- ### Exercise 6.8 (i)
|
||||
/-- ### Exercise 3.8 (i)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_6_8_i {A : Set (Set.Relation α)}
|
||||
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
|
||||
|
|
@ -355,14 +355,14 @@ theorem exercise_6_8_i {A : Set (Set.Relation α)}
|
|||
· intro ⟨t, ht, y, hx⟩
|
||||
exact ⟨y, t, ht, hx⟩
|
||||
|
||||
/-- ### Exercise 6.8 (ii)
|
||||
/-- ### Exercise 3.8 (ii)
|
||||
|
||||
Show that for any set `𝓐`:
|
||||
```
|
||||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_6_8_ii {A : Set (Set.Relation α)}
|
||||
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
|
||||
|
|
@ -380,7 +380,7 @@ theorem exercise_6_8_ii {A : Set (Set.Relation α)}
|
|||
· intro ⟨y, ⟨hy, ⟨t, ht⟩⟩⟩
|
||||
exact ⟨t, ⟨y, ⟨hy, ht⟩⟩⟩
|
||||
|
||||
/-- ### Exercise 6.9 (i)
|
||||
/-- ### Exercise 3.9 (i)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
|
|
@ -388,7 +388,7 @@ operation in the preceding problem.
|
|||
dom ⋃ A = ⋃ { dom R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_6_9_i {A : Set (Set.Relation α)}
|
||||
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
|
||||
|
|
@ -406,7 +406,7 @@ theorem exercise_6_9_i {A : Set (Set.Relation α)}
|
|||
intro _ y hy R hR
|
||||
exact ⟨y, hy R hR⟩
|
||||
|
||||
/-- ### Exercise 6.9 (ii)
|
||||
/-- ### Exercise 3.9 (ii)
|
||||
|
||||
Discuss the result of replacing the union operation by the intersection
|
||||
operation in the preceding problem.
|
||||
|
|
@ -414,7 +414,7 @@ operation in the preceding problem.
|
|||
ran ⋃ A = ⋃ { ran R | R ∈ 𝓐 }
|
||||
```
|
||||
-/
|
||||
theorem exercise_6_9_ii {A : Set (Set.Relation α)}
|
||||
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
|
||||
|
|
|
|||
Loading…
Reference in New Issue