Use swap theorems in pigeonhole principle proof.
Establish precedent for embedding LaTeX proof into code.fin-dom-ran
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@ -20,7 +20,8 @@ A ∩ B = B ∩ A
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-/
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theorem commutative_law_i (A B : Set α)
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: A ∪ B = B ∪ A := calc A ∪ B
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: A ∪ B = B ∪ A :=
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calc A ∪ B
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_ = { x | x ∈ A ∨ x ∈ B } := rfl
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_ = { x | x ∈ B ∨ x ∈ A } := by
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ext
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@ -23,8 +23,16 @@ If `x ∈ C` and `y ∈ C`, then `⟨x, y⟩ ∈ 𝒫 𝒫 C`.
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-/
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lemma lemma_3b {C : Set α} (hx : x ∈ C) (hy : y ∈ C)
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: OrderedPair x y ∈ 𝒫 𝒫 C := by
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/-
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> Let `C` be an arbitrary set and `x, y ∈ C`. Then by definition of the power
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> set, `{x}` and `{x, y}` are members of `𝒫 C`.
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-/
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have hxs : {x} ⊆ C := Set.singleton_subset_iff.mpr hx
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have hxys : {x, y} ⊆ C := Set.mem_mem_imp_pair_subset hx hy
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/-
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> Likewise `{{x}, {x, y}}` is a member of `𝒫 𝒫 C`. By definition of an ordered
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> pair, `⟨x, y⟩ = {{x}, {x, y}}`. This concludes our proof.
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-/
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exact Set.mem_mem_imp_pair_subset hxs hxys
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/-- #### Theorem 3D
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@ -33,10 +41,23 @@ If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to `⋃ ⋃ A`.
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-/
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theorem theorem_3d {A : Set (Set (Set α))} (h : OrderedPair x y ∈ A)
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: x ∈ ⋃₀ (⋃₀ A) ∧ y ∈ ⋃₀ (⋃₀ A) := by
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/-
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> Let `A` be a set and `⟨x, y⟩ ∈ A`. By definition of an ordered pair,
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>
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> `⟨x, y⟩ = {{x}, {x, y}}`.
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>
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> By Exercise 2.3, `{{x}, {x, y}} ⊆ ∪ A`. Then `{x, y} ∈ ∪ A`.
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-/
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have hp := Chapter_2.exercise_2_3 (OrderedPair x y) h
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unfold OrderedPair at hp
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have hq : {x, y} ∈ ⋃₀ A := hp (by simp)
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/-
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> Another application of Exercise 2.3 implies `{x, y} ∈ ∪ ∪ A`.
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-/
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have : {x, y} ⊆ ⋃₀ ⋃₀ A := Chapter_2.exercise_2_3 {x, y} hq
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/-
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> Therefore `x, y ∈ ∪ ∪ A`.
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-/
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exact ⟨this (by simp), this (by simp)⟩
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@ -72,28 +72,72 @@ lemma pigeonhole_principle_aux (n : ℕ)
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∀ f : ℕ → ℕ,
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Set.MapsTo f M (Set.Iio n) ∧ Set.InjOn f M →
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¬ Set.SurjOn f M (Set.Iio n) := by
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/-
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> Let
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>
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> `S = {n ∈ ω | ∀ M ⊂ n, every one-to-one function f: M → n is not onto}`. (1)
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>
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> We show that (i) `0 ∈ S` and (ii) if `n ∈ S`, then so is `n⁺`. Afterward we
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> prove (iii) the theorem statement.
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-/
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induction n with
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/-
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## (i)
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> By definition, `0 = ∅`.
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-/
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| zero =>
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intro _ hM
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unfold Set.Iio at hM
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simp only [Nat.zero_eq, not_lt_zero', Set.setOf_false] at hM
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/-
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> Then `0` has no proper subsets.
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-/
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rw [Set.ssubset_empty_iff_false] at hM
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/-
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> Hence `0 ∈ S` vacuously.
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-/
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exact False.elim hM
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/-
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## (ii)
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> Suppose `n ∈ S` and `M ⊂ n⁺`. Furthermore, let `f: M → n⁺` be a one-to-one
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> function.
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-/
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| succ n ih =>
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intro M hM f ⟨hf_maps, hf_inj⟩ hf_surj
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/-
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> If `M = ∅`, it vacuously holds that `f` is not onto `n⁺`.
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-/
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by_cases hM' : M = ∅
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· rw [hM', Set.SurjOn_emptyset_Iio_iff_eq_zero] at hf_surj
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simp at hf_surj
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/-
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> Otherwise `M ≠ 0`. Because `M` is finite, the trichotomy law for `ω` implies
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> the existence of a largest member `p ∈ M`. There are two cases to consider:
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-/
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by_cases h : ¬ ∃ t, t ∈ M ∧ f t = n
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-- Trivial case. `f` must not be onto if this is the case.
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/-
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### Case 1
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> `n ∉ ran f`.
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> Then `f` is not onto `n⁺`.
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-/
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· have ⟨t, ht⟩ := hf_surj (show n ∈ _ by simp)
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exact absurd ⟨t, ht⟩ h
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-- Continue under the assumption `n ∈ ran f`.
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/-
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### Case 2
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> `n ∈ ran f`.
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> Then there exists some `t ∈ M` such that `⟨t, n⟩ ∈ f`.
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-/
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have ⟨t, ht₁, ht₂⟩ := not_not.mp h
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/-
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> Define `f': M → n⁺` given by
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>
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> `f'(p) = f(t) = n`
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> `f'(t) = f(p)`
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> `f'(x) = f(x)` for all other `x`.
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>
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> That is, `f'` is a variant of `f` in which the largest element of its domain
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> (i.e. `p`) corresponds to value `n`.
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-/
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-- `M ≠ ∅` so `∃ p, ∀ x ∈ M, p ≥ x`, i.e. a maximum member.
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have ⟨p, hp₁, hp₂⟩ : ∃ p ∈ M, ∀ x, x ∈ M → p ≥ x := by
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refine subset_finite_max_nat (show Set.Finite M from ?_) ?_ ?_
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@ -104,69 +148,19 @@ lemma pigeonhole_principle_aux (n : ℕ)
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exact Set.nmem_singleton_empty.mp hM'
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· show M ⊆ M
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exact Eq.subset rfl
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-- `g` is a variant of `f` in which the largest element of its domain
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-- (i.e. `p`) corresponds to value `n`.
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let g x := if x = p then n else if x = t then f p else f x
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have hg_maps : Set.MapsTo g M (Set.Iio (n + 1)) := by
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intro x hx
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dsimp only
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by_cases hx₁ : x = p
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· rw [hx₁]
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simp
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· rw [if_neg hx₁]
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by_cases hx₂ : x = t
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· rw [hx₂]
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simp only [ite_true, Set.mem_Iio]
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exact hf_maps hp₁
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· rw [if_neg hx₂]
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simp only [Set.mem_Iio]
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exact hf_maps hx
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have hg_inj : Set.InjOn g M := by
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intro x₁ hx₁ x₂ hx₂ hf'
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by_cases hc₁ : x₁ = p
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· by_cases hc₂ : x₂ = p
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· rw [hc₁, hc₂]
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· dsimp at hf'
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rw [hc₁] at hf'
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simp only [ite_self, ite_true] at hf'
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by_cases hc₃ : x₂ = t
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· rw [if_neg hc₂, if_pos hc₃, ← ht₂] at hf'
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rw [hc₁] at hx₁ ⊢
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rw [hc₃] at hx₂ ⊢
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exact hf_inj hx₁ hx₂ hf'.symm
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· rw [if_neg hc₂, if_neg hc₃, ← ht₂] at hf'
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have := hf_inj ht₁ hx₂ hf'
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exact absurd this.symm hc₃
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· by_cases hc₂ : x₂ = p
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· rw [hc₂] at hf'
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simp only [ite_self, ite_true] at hf'
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by_cases hc₃ : x₁ = t
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· rw [if_neg hc₁, if_pos hc₃, ← ht₂] at hf'
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rw [hc₃] at hx₁ ⊢
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rw [hc₂] at hx₂ ⊢
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have := hf_inj hx₂ hx₁ hf'
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exact this.symm
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· rw [if_neg hc₁, if_neg hc₃, ← ht₂] at hf'
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have := hf_inj hx₁ ht₁ hf'
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exact absurd this hc₃
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· dsimp only at hf'
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rw [if_neg hc₁, if_neg hc₂] at hf'
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by_cases hc₃ : x₁ = t
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· by_cases hc₄ : x₂ = t
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· rw [hc₃, hc₄]
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· rw [if_pos hc₃, if_neg hc₄] at hf'
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have := hf_inj hp₁ hx₂ hf'
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exact absurd this.symm hc₂
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· by_cases hc₄ : x₂ = t
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· rw [if_neg hc₃, if_pos hc₄] at hf'
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have := hf_inj hx₁ hp₁ hf'
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exact absurd this hc₁
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· rw [if_neg hc₃, if_neg hc₄] at hf'
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exact hf_inj hx₁ hx₂ hf'
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/-
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> Next define `g = f' - {⟨p, n⟩}`. Then `g` is a function mapping `M - {p}` to
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> `n`.
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-/
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let g := Set.Function.swap f p t
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/-
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> Since `f` is one-to-one, `f'` and `g` are also one-to-one.
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-/
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have hg_maps := Set.Function.swap_MapsTo_self hp₁ ht₁ hf_maps
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have hg_inj := Set.Function.swap_InjOn_self hp₁ ht₁ hf_inj
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/-
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> Then (1) indicates `g` must not be onto `n`.
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-/
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let M' := M \ {p}
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have hM' : M' ⊂ Set.Iio n := by
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by_cases hc : p = n
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@ -209,18 +203,28 @@ lemma pigeonhole_principle_aux (n : ℕ)
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· -- `Set.MapsTo g M' (Set.Iio n)`
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intro x hx
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have hx₁ : x ∈ M := Set.mem_of_mem_diff hx
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apply Or.elim (Nat.lt_or_eq_of_lt $ hg_maps hx₁)
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· exact id
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· intro hx₂
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rw [← show g p = n by simp] at hx₂
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exact absurd (hg_inj hx₁ hp₁ hx₂) hx.right
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apply Or.elim (Nat.lt_or_eq_of_lt $ hg_maps hx₁) id
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intro hx₂
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unfold Set.Function.swap at hx₂
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by_cases hc₁ : x = p
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· exact absurd hc₁ hx.right
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· rw [if_neg hc₁] at hx₂
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by_cases hc₂ : x = t
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· rw [if_pos hc₂, ← ht₂] at hx₂
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have := hf_inj hp₁ ht₁ hx₂
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rw [← hc₂] at this
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exact absurd this.symm hc₁
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· rw [if_neg hc₂, ← ht₂] at hx₂
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have := hf_inj hx₁ ht₁ hx₂
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exact absurd this hc₂
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· -- `Set.InjOn g M'`
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intro x₁ hx₁ x₂ hx₂ hg
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have hx₁' : x₁ ∈ M := (Set.diff_subset M {p}) hx₁
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have hx₂' : x₂ ∈ M := (Set.diff_subset M {p}) hx₂
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exact hg_inj hx₁' hx₂' hg
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-- We have shown `g` isn't surjective. This is another way of saying that.
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/-
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> That is, there exists some `a ∈ n` such that `a ∉ ran g`.
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-/
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have ⟨a, ha₁, ha₂⟩ : ∃ a, a < n ∧ a ∉ g '' M' := by
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unfold Set.SurjOn at ng_surj
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rw [Set.subset_def] at ng_surj
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@ -235,8 +239,11 @@ lemma pigeonhole_principle_aux (n : ℕ)
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unfold Set.image
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simp only [Set.mem_Iio, Set.mem_setOf_eq, not_exists, not_and]
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exact ng_surj
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-- If `g` isn't surjective then neither is `f`.
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/-
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> By the trichotomy law for `ω`, `a ≠ n`. Therefore `a ∉ ran f'`.
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> `ran f' = ran f` meaning `a ∉ ran f`. Because `a ∈ n ∈n⁺`, Theorem 4F implies
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> `a ∈ n⁺`. Hence `f` is not onto `n⁺`.
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-/
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refine absurd (hf_surj $ calc a
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_ < n := ha₁
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_ < n + 1 := by simp) (show ↑a ∉ f '' M from ?_)
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@ -248,17 +255,20 @@ lemma pigeonhole_principle_aux (n : ℕ)
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simp only [Set.mem_Iio, Set.mem_setOf_eq, not_exists, not_and] at ha₂ ⊢
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intro x hx
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by_cases hxp : x = p
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· rw [if_pos hxp]
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· unfold Set.Function.swap
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rw [if_pos hxp, ht₂]
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exact (Nat.ne_of_lt ha₁).symm
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· refine ha₂ x ?_
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exact Set.mem_diff_of_mem hx hxp
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ext x
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dsimp only
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unfold Set.Function.swap
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simp only [Set.mem_image, Set.mem_Iio]
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apply Iff.intro
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· intro ⟨y, hy₁, hy₂⟩
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by_cases hc₁ : y = p
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· rw [if_pos hc₁] at hy₂
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· rw [if_pos hc₁, ht₂] at hy₂
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rw [hy₂] at ht₂
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exact ⟨t, ht₁, ht₂⟩
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· rw [if_neg hc₁] at hy₂
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@ -274,15 +284,25 @@ lemma pigeonhole_principle_aux (n : ℕ)
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· rw [hc₂, ht₂] at hy₂
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rw [← hc₁, ← hc₂]
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simp only [ite_self, ite_true]
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exact hy₂
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rwa [hc₂, ht₂]
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· rw [hc₁, ← Ne.def] at hc₂
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rwa [if_neg hc₂.symm, if_pos rfl, ← hc₁]
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· by_cases hc₂ : y = t
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· refine ⟨p, hp₁, ?_⟩
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simp only [ite_self, ite_true]
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rwa [hc₂, ht₂] at hy₂
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rwa [hc₂] at hy₂
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· refine ⟨y, hy₁, ?_⟩
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rwa [if_neg hc₁, if_neg hc₂]
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/-
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### Subconclusion
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> The foregoing cases are exhaustive. Hence `n⁺ ∈ S`.
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## (iii)
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> By (i) and (ii), `S` is an inductive set. By Theorem 4B, `S = ω`. Thus for all
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> natural numbers `n`, there is no one-to-one correspondence between `n` and a
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> proper subset of `n`. In other words, no natural number is equinumerous to a
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> proper subset of itself.
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-/
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/--
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No natural number is equinumerous to a proper subset of itself.
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@ -5,7 +5,7 @@ import Mathlib.Data.Set.Function
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Additional theorems around functions defined on sets.
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-/
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namespace Set
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namespace Set.Function
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/--
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Produce a new function that swaps the outputs of the two specified inputs.
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@ -232,4 +232,4 @@ theorem self_iff_swap_BijOn [DecidableEq α]
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: BijOn (swap f a₁ a₂) A B ↔ BijOn f A B :=
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⟨self_BijOn_swap ha₁ ha₂, swap_BijOn_self ha₁ ha₂⟩
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end Set
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end Set.Function
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