Enderton. Formally verify exercises 3.12 through 3.15.
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@ -3949,7 +3949,7 @@ Then $F = G$.
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\end{proof}
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\subsection{\pending{Exercise 3.12}}%
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\subsection{\verified{Exercise 3.12}}%
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\label{sub:exercise-3.12}
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Assume that $f$ and $g$ are functions and show that
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@ -3958,6 +3958,9 @@ Assume that $f$ and $g$ are functions and show that
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_12}
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Let $f$ and $g$ be \nameref{ref:function}s.
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\paragraph{($\Rightarrow$)}%
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@ -3982,7 +3985,7 @@ Assume that $f$ and $g$ are functions and show that
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\end{proof}
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\subsection{\pending{Exercise 3.13}}%
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\subsection{\verified{Exercise 3.13}}%
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\label{sub:exercise-3.13}
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Assume that $f$ and $g$ are functions with $f \subseteq g$ and
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@ -3991,6 +3994,9 @@ Show that $f = g$.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_13}
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Let $f$ and $g$ be functions such that $f \subseteq g$ and
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$\dom{g} \subseteq \dom{f}$.
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By \nameref{sub:exercise-3.12}, it follows that $\dom{f} \subseteq \dom{g}$
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@ -4001,7 +4007,7 @@ Show that $f = g$.
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\end{proof}
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\subsection{\pending{Exercise 3.14}}%
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\subsection{\verified{Exercise 3.14}}%
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\label{sub:exercise-3.14}
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Assume that $f$ and $g$ are functions.
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@ -4014,17 +4020,22 @@ Assume that $f$ and $g$ are functions.
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\begin{proof}
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Assume $f$ and $g$ are functions.
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_14\_a}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_14\_b}
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Assume $f$ and $g$ are \nameref{ref:function}s.
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\paragraph{(a)}%
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Consider $f \cap g$.
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By definition of the intersection of sets, $f \cap g \subseteq f$.
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By \nameref{sub:exercise-3.12}, $\dom{(f \cap g)} = \dom{f}$ and
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$(\forall x \in \dom{(f \cap g)} (f \cap g)(x) = f(x)$.
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The latter conjunct shows that, since $f$ is single-valued, $f \cap g$ must
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also be single-valued.
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In other words, $f \cap g$ is a function.
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Since $f$ is single-valued, it trivially follows that so must $f \cap g$.
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Therefore $f \cap g$ is a function.
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\paragraph{(b)}%
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@ -4064,7 +4075,7 @@ Assume that $f$ and $g$ are functions.
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\end{proof}
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\subsection{\pending{Exercise 3.15}}%
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\subsection{\verified{Exercise 3.15}}%
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\label{sub:exercise-3.15}
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Let $\mathscr{A}$ be a set of functions such that for any $f$ and $g$ in
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@ -4073,6 +4084,9 @@ Show that $\bigcup{\mathscr{A}}$ is a function.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_15}
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Let $\mathscr{A}$ be a set of \nameref{ref:function}s such that for any $f$
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and $g$ in $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$.
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Let $x \in \dom{\bigcup{\mathscr{A}}}$.
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@ -325,7 +325,7 @@ theorem exercise_3_7 {R : Set.Relation α}
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simp only [Set.mem_singleton_iff, Set.mem_insert_iff] at this
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exact hxy_mem this
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section
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section Relation
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open Set.Relation
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@ -678,11 +678,8 @@ theorem theorem_3k_c_ii {F : Set.Relation α} {A B : Set α}
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unfold image at nv
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simp only [Set.mem_setOf_eq] at nv
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have ⟨u₁, hu₁⟩ := nv
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have ⟨x, hx⟩ := hF v (mem_pair_imp_snd_mem_ran hu.right)
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simp only [and_imp] at hx
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have hr₁ := hx.right u (mem_pair_imp_fst_mem_dom hu.right) hu.right
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have hr₂ := hx.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
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rw [hr₂, ← hr₁] at hu₁
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have := single_rooted_eq_unique hF hu.right hu₁.right
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rw [← this] at hu₁
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exact absurd hu₁.left hu.left.right
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exact ⟨hv₁, hv₂⟩
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@ -714,6 +711,173 @@ theorem corollary_3l_iii {G : Set.Relation α} {A B : Set α}
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single_valued_self_iff_single_rooted_inv.mp hG
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exact (theorem_3k_c_ii hG').symm
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end
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/-- #### Exercise 3.12
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Assume that `f` and `g` are functions and show that
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```
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f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f) f(x) = g(x).
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```
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-/
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theorem exercise_3_12 {f g : Set.Relation α}
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(hf : f.isSingleValued) (_ : g.isSingleValued)
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: f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f, ∃! y : α, (x, y) ∈ f ∧ (x, y) ∈ g) := by
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apply Iff.intro
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· intro h
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apply And.intro
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· show ∀ x, x ∈ dom f → x ∈ dom g
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intro x hx
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have ⟨y, hy⟩ := dom_exists hx
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exact mem_pair_imp_fst_mem_dom (h hy)
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· intro x hx
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have ⟨y, hy⟩ := dom_exists hx
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refine ⟨y, ⟨hy, h hy⟩, ?_⟩
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intro y₁ hy₁
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exact single_valued_eq_unique hf hy₁.left hy
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· intro ⟨_, hx⟩
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show ∀ p, p ∈ f → p ∈ g
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intro (x, y) hp
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have ⟨y₁, hy₁⟩ := hx x (mem_pair_imp_fst_mem_dom hp)
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rw [single_valued_eq_unique hf hp hy₁.left.left]
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exact hy₁.left.right
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/-- #### Exercise 3.13
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Assume that `f` and `g` are functions with `f ⊆ g` and `dom g ⊆ dom f`. Show
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that `f = g`.
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-/
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theorem exercise_3_13 {f g : Set.Relation α}
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(hf : f.isSingleValued) (hg : g.isSingleValued)
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(h : f ⊆ g) (h₁ : dom g ⊆ dom f)
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: f = g := by
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have h₂ := (exercise_3_12 hf hg).mp h
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have hfg := Set.Subset.antisymm_iff.mpr ⟨h₁, h₂.left⟩
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ext p
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have (a, b) := p
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apply Iff.intro
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· intro hp
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have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp)
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rw [single_valued_eq_unique hf hp hx.left.left]
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exact hx.left.right
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· intro hp
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rw [← hfg] at h₂
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have ⟨x, hx⟩ := h₂.right a (mem_pair_imp_fst_mem_dom hp)
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rw [single_valued_eq_unique hg hp hx.left.right]
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exact hx.left.left
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/-- #### Exercise 3.14 (a)
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Assume that `f` and `g` are functions. Show that `f ∩ g` is a function.
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-/
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theorem exercise_3_14_a {f g : Set.Relation α}
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(hf : f.isSingleValued) (_ : g.isSingleValued)
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: (f ∩ g).isSingleValued :=
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single_valued_subset hf (Set.inter_subset_left f g)
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/-- #### Exercise 3.14 (b)
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Assume that `f` and `g` are functions. Show that `f ∪ g` is a function **iff**
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`f(x) = g(x)` for every `x` in `(dom f) ∩ (dom g)`.
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-/
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theorem exercise_3_14_b {f g : Set.Relation α}
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(hf : f.isSingleValued) (hg : g.isSingleValued)
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: (f ∪ g).isSingleValued ↔
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(∀ x ∈ dom f ∩ dom g, ∃! y, (x, y) ∈ f ∧ (x, y) ∈ g) := by
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apply Iff.intro
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· intro h x hx
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have ⟨y₁, hy₁⟩ := hf x hx.left
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have ⟨y₂, hy₂⟩ := hg x hx.right
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have : y₁ = y₂ := by
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have hf' : (x, y₁) ∈ f ∪ g := Or.inl hy₁.left.right
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have hg' : (x, y₂) ∈ f ∪ g := Or.inr hy₂.left.right
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exact single_valued_eq_unique h hf' hg'
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rw [← this] at hy₂
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refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩
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intro y₃ hfy₃
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exact single_valued_eq_unique hf hfy₃.left hy₁.left.right
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· intro h x hx
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by_cases hfx : x ∈ dom f
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· by_cases hgx : x ∈ dom g
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· -- `x ∈ dom f ∧ x ∈ dom g`
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have ⟨y₁, hy₁⟩ := hf x hfx
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have ⟨y₂, hy₂⟩ := hg x hgx
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refine ⟨y₁, ⟨?_, Or.inl hy₁.left.right⟩, ?_⟩
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· unfold ran
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simp only [Set.mem_image, Set.mem_union, Prod.exists, exists_eq_right]
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exact ⟨x, Or.inl hy₁.left.right⟩
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· intro y₃ hy₃
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apply Or.elim hy₃.right
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· intro hxy
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exact single_valued_eq_unique hf hxy hy₁.left.right
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· refine fun hxy => single_valued_eq_unique hg hxy ?_
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have : y₁ = y₂ := by
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have ⟨z, ⟨hz, _⟩⟩ := h x ⟨hfx, hgx⟩
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rw [
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single_valued_eq_unique hf hy₁.left.right hz.left,
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single_valued_eq_unique hg hy₂.left.right hz.right
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]
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rw [this]
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exact hy₂.left.right
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· -- `x ∈ dom f ∧ x ∉ dom g`
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have ⟨y, hy⟩ := dom_exists hfx
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have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_left f g) hy
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refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
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intro y₁ hy₁
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apply Or.elim hy₁.right
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· intro hx'
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exact single_valued_eq_unique hf hx' hy
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· intro hx'
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exact absurd (mem_pair_imp_fst_mem_dom hx') hgx
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· by_cases hgx : x ∈ dom g
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· -- `x ∉ dom f ∧ x ∈ dom g`
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have ⟨y, hy⟩ := dom_exists hgx
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have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_right f g) hy
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refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
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intro y₁ hy₁
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apply Or.elim hy₁.right
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· intro hx'
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exact absurd (mem_pair_imp_fst_mem_dom hx') hfx
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· intro hx'
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exact single_valued_eq_unique hg hx' hy
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· -- `x ∉ dom f ∧ x ∉ dom g`
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exfalso
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unfold dom at hx
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simp only [
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Set.mem_image,
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Set.mem_union,
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Prod.exists,
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exists_and_right,
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exists_eq_right
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] at hx
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have ⟨_, hy⟩ := hx
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apply Or.elim hy
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· intro hz
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exact absurd (mem_pair_imp_fst_mem_dom hz) hfx
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· intro hz
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exact absurd (mem_pair_imp_fst_mem_dom hz) hgx
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/-- #### Exercise 3.15
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Let `𝓐` be a set of functions such that for any `f` and `g` in `𝓐`, either
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`f ⊆ g` or `g ⊆ f`. Show that `⋃ 𝓐` is a function.
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-/
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theorem exercise_3_15 {𝓐 : Set (Set.Relation α)}
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(h𝓐 : ∀ F ∈ 𝓐, F.isSingleValued)
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(h : ∀ F, ∀ G, F ∈ 𝓐 → G ∈ 𝓐 → F ⊆ G ∨ G ⊆ F)
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: isSingleValued (⋃₀ 𝓐) := by
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intro x hx
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have ⟨y₁, hy₁⟩ := dom_exists hx
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refine ⟨y₁, ⟨mem_pair_imp_snd_mem_ran hy₁, hy₁⟩, ?_⟩
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intro y₂ hy₂
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have ⟨f, hf⟩ : ∃ f : Set.Relation α, f ∈ 𝓐 ∧ (x, y₁) ∈ f := hy₁
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have ⟨g, hg⟩ : ∃ g : Set.Relation α, g ∈ 𝓐 ∧ (x, y₂) ∈ g := hy₂.right
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apply Or.elim (h f g hf.left hg.left)
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· intro hf'
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have := hf' hf.right
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exact single_valued_eq_unique (h𝓐 g hg.left) hg.right this
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· intro hg'
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have := hg' hg.right
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exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right
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end Relation
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end Enderton.Set.Chapter_3
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@ -180,7 +180,7 @@ A `Relation` `R` is said to be single-rooted **iff** for all `y ∈ ran R`, ther
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exists exactly one `x` such that `⟨x, y⟩ ∈ R`.
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-/
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def isSingleRooted (R : Relation α) : Prop :=
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∀ y ∈ R.ran, ∃! x, x ∈ R.dom ∧ (x, y) ∈ R
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∀ y ∈ ran R, ∃! x, x ∈ dom R ∧ (x, y) ∈ R
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/--
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A single-rooted `Relation` should map the same output to the same input.
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@ -204,7 +204,7 @@ exists exactly one `y` such that `⟨x, y⟩ ∈ R`.
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Notice, a `Relation` that is single-valued is a function.
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-/
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def isSingleValued (R : Relation α) : Prop :=
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∀ x ∈ R.dom, ∃! y, y ∈ R.ran ∧ (x, y) ∈ R
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∀ x ∈ dom R, ∃! y, y ∈ ran R ∧ (x, y) ∈ R
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/--
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A single-valued `Relation` should map the same input to the same output.
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@ -263,6 +263,21 @@ theorem single_valued_self_iff_single_rooted_inv {F : Set.Relation α}
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conv => lhs; rw [← inv_inv_eq_self F]
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rw [single_valued_inv_iff_single_rooted_self]
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/--
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The subset of a function must also be a function.
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-/
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theorem single_valued_subset {F G : Set.Relation α}
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(hG : G.isSingleValued) (h : F ⊆ G)
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: F.isSingleValued := by
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unfold isSingleValued
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intro x hx
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have ⟨y, hy⟩ := dom_exists hx
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unfold ExistsUnique
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simp only
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refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hy, hy⟩, ?_⟩
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intro y₁ hy₁
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exact single_valued_eq_unique hG (h hy₁.right) (h hy)
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/--
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A `Relation` `R` is one-to-one if it is a single-rooted function.
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-/
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