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fix: label Ex. 11.6.4 (a)/(b) and Ex. 3.1.7–3.1.9 parts #617

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fix: label Ex. 11.6.4 (a)/(b) and Ex. 3.1.7–3.1.9 parts #617

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3 changes: 2 additions & 1 deletion Analysis/Section_11_6.lean
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Expand Up @@ -154,10 +154,11 @@ theorem summable_iff_integ_of_antitone {f:ℝ → ℝ} (hnon: ∀ x ≥ 0, f x
-- Exercise 11.6.2: Formulate a reasonable notion of a piecewise monotone function, and then
-- show that all bounded piecewise monotone functions are Riemann integrable.

/-- Exercise 11.6.4 -/
/-- Exercise 11.6.4 (a) -/
example : ∃ (f:ℝ → ℝ), (∀ x ≥ 0, f x ≥ 0) ∧ Summable (fun n:ℕ ↦ f n) ∧ ¬ ∃ M, ∀ N ≥ 0, integ f (Icc 0 N) ≤ M := by
sorry

/-- Exercise 11.6.4 (b) -/
example : ∃ (f:ℝ → ℝ), (∀ x ≥ 0, f x ≥ 0) ∧ ¬ Summable (fun n:ℕ ↦ f n) ∧ ∃ M, ∀ N ≥ 0, integ f (Icc 0 N) ≤ M := by
sorry

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20 changes: 10 additions & 10 deletions Analysis/Section_3_1.lean
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Expand Up @@ -731,45 +731,45 @@ example : ({3,5,9}:Set).replace (P := fun _ y ↦ y=1) (by aesop) = {1} := by
/-- Exercise 3.1.5. One can use the {tactic}`tfae_have` and {tactic}`tfae_finish` tactics here. -/
theorem SetTheory.Set.subset_tfae (A B:Set) : [A ⊆ B, A ∪ B = B, A ∩ B = A].TFAE := by sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (a) -/
theorem SetTheory.Set.inter_subset_left (A B:Set) : A ∩ B ⊆ A := by
sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (b) -/
theorem SetTheory.Set.inter_subset_right (A B:Set) : A ∩ B ⊆ B := by
sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (c) -/
@[simp]
theorem SetTheory.Set.subset_inter_iff (A B C:Set) : C ⊆ A ∩ B ↔ C ⊆ A ∧ C ⊆ B := by
sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (d) -/
theorem SetTheory.Set.subset_union_left (A B:Set) : A ⊆ A ∪ B := by
sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (e) -/
theorem SetTheory.Set.subset_union_right (A B:Set) : B ⊆ A ∪ B := by
sorry

/-- Exercise 3.1.7 -/
/-- Exercise 3.1.7 (f) -/
@[simp]
theorem SetTheory.Set.union_subset_iff (A B C:Set) : A ∪ B ⊆ C ↔ A ⊆ C ∧ B ⊆ C := by
sorry

/-- Exercise 3.1.8 -/
/-- Exercise 3.1.8 (a) -/
@[simp]
theorem SetTheory.Set.inter_union_cancel (A B:Set) : A ∩ (A ∪ B) = A := by sorry

/-- Exercise 3.1.8 -/
/-- Exercise 3.1.8 (b) -/
@[simp]
theorem SetTheory.Set.union_inter_cancel (A B:Set) : A ∪ (A ∩ B) = A := by sorry

/-- Exercise 3.1.9 -/
/-- Exercise 3.1.9 (a) -/
theorem SetTheory.Set.partition_left {A B X:Set} (h_union: A ∪ B = X) (h_inter: A ∩ B = ∅) :
A = X \ B := by sorry

/-- Exercise 3.1.9 -/
/-- Exercise 3.1.9 (b) -/
theorem SetTheory.Set.partition_right {A B X:Set} (h_union: A ∪ B = X) (h_inter: A ∩ B = ∅) :
B = X \ A := by
sorry
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