86 lines
2.0 KiB
Plaintext
86 lines
2.0 KiB
Plaintext
import Mathlib.Data.Set.Function
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import Mathlib.Data.Rel
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/-! # Enderton.Set.Chapter_6
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Cardinal Numbers and the Axiom of Choice
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-/
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namespace Enderton.Set.Chapter_6
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/-! #### Theorem 6A
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For any sets `A`, `B`, and `C`,
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(a) `A ≈ A`.
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(b) If `A ≈ B`, then `B ≈ A`.
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(c) If `A ≈ B` and `B ≈ C`, then `A ≈ C`.
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-/
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theorem theorem_6a_a (A : Set α)
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: ∃ F, Set.BijOn F A A := by
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refine ⟨fun x => x, ?_⟩
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unfold Set.BijOn Set.MapsTo Set.InjOn Set.SurjOn
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simp only [imp_self, implies_true, Set.image_id', true_and]
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exact Eq.subset rfl
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theorem theorem_6a_b [Nonempty α] (A : Set α) (B : Set β)
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(F : α → β) (hF : Set.BijOn F A B)
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: ∃ G, Set.BijOn G B A := by
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refine ⟨Function.invFunOn F A, ?_⟩
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exact (Set.bijOn_comm $ Set.BijOn.invOn_invFunOn hF).mpr hF
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theorem theorem_6a_c (A : Set α) (B : Set β) (C : Set γ)
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(F : α → β) (hF : Set.BijOn F A B)
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(G : β → γ) (hG : Set.BijOn G B C)
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: ∃ H, Set.BijOn H A C := by
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exact ⟨G ∘ F, Set.BijOn.comp hG hF⟩
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/-- #### Exercise 6.1
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Show that the equation
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```
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f(m, n) = 2ᵐ(2n + 1) - 1
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```
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defines a one-to-one correspondence between `ω × ω` and `ω`.
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-/
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theorem exercise_6_1
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: Function.Bijective (fun p : ℕ × ℕ => 2 ^ p.1 * (2 * p.2 + 1) - 1) := by
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sorry
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/-- #### Exercise 6.2
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Show that in Fig. 32 we have:
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```
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J(m, n) = [1 + 2 + ⋯ + (m + n)] + m
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= (1 / 2)[(m + n)² + 3m + n].
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```
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-/
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theorem exercise_6_2
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: Function.Bijective
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(fun p : ℕ × ℕ => (1 / 2) * ((p.1 + p.2) ^ 2 + 3 * p.1 + p.2)) := by
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sorry
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/-- #### Exercise 6.3
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Find a one-to-one correspondence between the open unit interval `(0, 1)` and `ℝ`
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that takes rationals to rationals and irrationals to irrationals.
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-/
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theorem exercise_6_3
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: True := by
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sorry
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/-- #### Exercise 6.4
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Construct a one-to-one correspondence between the closed unit interval
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```
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[0, 1] = {x ∈ ℝ | 0 ≤ x ≤ 1}
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```
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and the open unit interval `(0, 1)`.
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-/
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theorem exercise_6_4
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: ∃ F, Set.BijOn F (Set.Ioo 0 1) (@Set.univ ℝ) := by
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sorry
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end Enderton.Set.Chapter_6
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