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Enderton. Exercises 3.16 - 3.18.

finite-set-exercises
Joshua Potter 2023-07-05 16:04:43 -06:00
parent d3f242cf07
commit 0e55484adc
2 changed files with 260 additions and 68 deletions
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30
Bookshelf/Enderton/Set.tex
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@ -4137,7 +4137,7 @@ Show that there is no set to which every function belongs.
\end{proof} \end{proof}
\subsection{\pending{Exercise 3.17}}% \subsection{\verified{Exercise 3.17}}%
\label{sub:exercise-3.17} \label{sub:exercise-3.17}
Show that the composition of two single-rooted sets is again single-rooted. Show that the composition of two single-rooted sets is again single-rooted.
@ -4145,6 +4145,14 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
\begin{proof} \begin{proof}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_17\_i}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_17\_ii}
Let $F$ and $G$ be two single-rooted sets. Let $F$ and $G$ be two single-rooted sets.
Consider $F \circ G$. Consider $F \circ G$.
By definition of the \nameref{ref:composition} of sets, By definition of the \nameref{ref:composition} of sets,
@ -4174,7 +4182,7 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
\end{proof} \end{proof}
\subsection{\pending{Exercise 3.18}}% \subsection{\verified{Exercise 3.18}}%
\label{sub:exercise-3.18} \label{sub:exercise-3.18}
Let $R$ be the set Let $R$ be the set
@ -4185,11 +4193,27 @@ Evaluate the following: $R \circ R$, $R \restriction \{1\}$,
\begin{proof} \begin{proof}
\statementpadding
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_18\_i}
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_18\_ii}
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_18\_iii}
\lean*{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_18\_iv}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_3\_18\_v}
\begin{enumerate}[(i)] \begin{enumerate}[(i)]
\item $R \circ R = \{ \item $R \circ R = \{
\left< 0, 2 \right>, \left< 0, 2 \right>,
\left< 0, 3 \right>, \left< 0, 3 \right>,
\left< 0, 3 \right>,
\left< 1, 3 \right> \left< 1, 3 \right>
\}$. \}$.
\item $R \restriction \{1\} = \{ \item $R \restriction \{1\} = \{

298
Bookshelf/Enderton/Set/Chapter_3.lean
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@ -1,6 +1,7 @@
import Bookshelf.Enderton.Set.Chapter_2 import Bookshelf.Enderton.Set.Chapter_2
import Bookshelf.Enderton.Set.OrderedPair import Bookshelf.Enderton.Set.OrderedPair
import Bookshelf.Enderton.Set.Relation import Bookshelf.Enderton.Set.Relation
import Mathlib.Tactic.CasesM
/-! # Enderton.Set.Chapter_3 /-! # Enderton.Set.Chapter_3
@ -469,7 +470,7 @@ dom (F ∘ G) = {x ∈ dom G | G(x) ∈ dom F}.
-/ -/
theorem theorem_3h_dom {F G : Set.Relation α} theorem theorem_3h_dom {F G : Set.Relation α}
(_ : F.isSingleValued) (hG : G.isSingleValued) (_ : F.isSingleValued) (hG : G.isSingleValued)
: dom (F.comp G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F } := by : dom (F.comp G) = {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F} := by
let rhs := {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F } let rhs := {x ∈ dom G | ∃! y, (x, y) ∈ G ∧ y ∈ dom F }
rw [Set.Subset.antisymm_iff] rw [Set.Subset.antisymm_iff]
apply And.intro apply And.intro
@ -720,7 +721,8 @@ f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f) f(x) = g(x).
-/ -/
theorem exercise_3_12 {f g : Set.Relation α} theorem exercise_3_12 {f g : Set.Relation α}
(hf : f.isSingleValued) (_ : g.isSingleValued) (hf : f.isSingleValued) (_ : g.isSingleValued)
: f ⊆ g ↔ dom f ⊆ dom g ∧ (∀ x ∈ dom f, ∃! y : α, (x, y) ∈ f ∧ (x, y) ∈ g) := by : f ⊆ g ↔ dom f ⊆ dom g ∧
(∀ x ∈ dom f, ∃! y : α, (x, y) ∈ f ∧ (x, y) ∈ g) := by
apply Iff.intro apply Iff.intro
· intro h · intro h
apply And.intro apply And.intro
@ -786,74 +788,72 @@ theorem exercise_3_14_b {f g : Set.Relation α}
· intro h x hx · intro h x hx
have ⟨y₁, hy₁⟩ := hf x hx.left have ⟨y₁, hy₁⟩ := hf x hx.left
have ⟨y₂, hy₂⟩ := hg x hx.right have ⟨y₂, hy₂⟩ := hg x hx.right
have : y₁ = y₂ := by have : y₁ = y₂ := single_valued_eq_unique h
have hf' : (x, y₁) ∈ f ∪ g := Or.inl hy₁.left.right (Or.inl hy₁.left.right)
have hg' : (x, y₂) ∈ f ∪ g := Or.inr hy₂.left.right (Or.inr hy₂.left.right)
exact single_valued_eq_unique h hf' hg'
rw [← this] at hy₂ rw [← this] at hy₂
refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩ refine ⟨y₁, ⟨hy₁.left.right, hy₂.left.right⟩, ?_⟩
intro y₃ hfy₃ intro y₃ hfy₃
exact single_valued_eq_unique hf hfy₃.left hy₁.left.right exact single_valued_eq_unique hf hfy₃.left hy₁.left.right
· intro h x hx · intro h x hx
by_cases hfx : x ∈ dom f by_cases hfx : x ∈ dom f <;>
· by_cases hgx : x ∈ dom g by_cases hgx : x ∈ dom g
· -- `x ∈ dom f ∧ x ∈ dom g` · -- `x ∈ dom f ∧ x ∈ dom g`
have ⟨y₁, hy₁⟩ := hf x hfx have ⟨y₁, hy₁⟩ := hf x hfx
have ⟨y₂, hy₂⟩ := hg x hgx have ⟨y₂, hy₂⟩ := hg x hgx
refine ⟨y₁, ⟨?_, Or.inl hy₁.left.right⟩, ?_⟩ refine ⟨y₁, ⟨?_, Or.inl hy₁.left.right⟩, ?_⟩
· unfold ran · unfold ran
simp only [Set.mem_image, Set.mem_union, Prod.exists, exists_eq_right] simp only [Set.mem_image, Set.mem_union, Prod.exists, exists_eq_right]
exact ⟨x, Or.inl hy₁.left.right⟩ exact ⟨x, Or.inl hy₁.left.right⟩
· intro y₃ hy₃ · intro y₃ hy₃
apply Or.elim hy₃.right apply Or.elim hy₃.right
· intro hxy · intro hxy
exact single_valued_eq_unique hf hxy hy₁.left.right exact single_valued_eq_unique hf hxy hy₁.left.right
· refine fun hxy => single_valued_eq_unique hg hxy ?_ · refine fun hxy => single_valued_eq_unique hg hxy ?_
have : y₁ = y₂ := by have : y₁ = y₂ := by
have ⟨z, ⟨hz, _⟩⟩ := h x ⟨hfx, hgx⟩ have ⟨z, ⟨hz, _⟩⟩ := h x ⟨hfx, hgx⟩
rw [ rw [
single_valued_eq_unique hf hy₁.left.right hz.left, single_valued_eq_unique hf hy₁.left.right hz.left,
single_valued_eq_unique hg hy₂.left.right hz.right single_valued_eq_unique hg hy₂.left.right hz.right
] ]
rw [this] rw [this]
exact hy₂.left.right exact hy₂.left.right
· -- `x ∈ dom f ∧ x ∉ dom g` · -- `x ∈ dom f ∧ x ∉ dom g`
have ⟨y, hy⟩ := dom_exists hfx have ⟨y, hy⟩ := dom_exists hfx
have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_left f g) hy have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_left f g) hy
refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩ refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
intro y₁ hy₁ intro y₁ hy₁
apply Or.elim hy₁.right apply Or.elim hy₁.right
· intro hx' · intro hx'
exact single_valued_eq_unique hf hx' hy exact single_valued_eq_unique hf hx' hy
· intro hx' · intro hx'
exact absurd (mem_pair_imp_fst_mem_dom hx') hgx exact absurd (mem_pair_imp_fst_mem_dom hx') hgx
· by_cases hgx : x ∈ dom g · -- `x ∉ dom f ∧ x ∈ dom g`
· -- `x ∉ dom f ∧ x ∈ dom g` have ⟨y, hy⟩ := dom_exists hgx
have ⟨y, hy⟩ := dom_exists hgx have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_right f g) hy
have hxy : (x, y) ∈ f ∪ g := (Set.subset_union_right f g) hy refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩
refine ⟨y, ⟨mem_pair_imp_snd_mem_ran hxy, hxy⟩, ?_⟩ intro y₁ hy₁
intro y₁ hy₁ apply Or.elim hy₁.right
apply Or.elim hy₁.right · intro hx'
· intro hx' exact absurd (mem_pair_imp_fst_mem_dom hx') hfx
exact absurd (mem_pair_imp_fst_mem_dom hx') hfx · intro hx'
· intro hx' exact single_valued_eq_unique hg hx' hy
exact single_valued_eq_unique hg hx' hy · -- `x ∉ dom f ∧ x ∉ dom g`
· -- `x ∉ dom f ∧ x ∉ dom g` exfalso
exfalso unfold dom at hx
unfold dom at hx simp only [
simp only [ Set.mem_image,
Set.mem_image, Set.mem_union,
Set.mem_union, Prod.exists,
Prod.exists, exists_and_right,
exists_and_right, exists_eq_right
exists_eq_right ] at hx
] at hx have ⟨_, hy⟩ := hx
have ⟨_, hy⟩ := hx apply Or.elim hy
apply Or.elim hy · intro hz
· intro hz exact absurd (mem_pair_imp_fst_mem_dom hz) hfx
exact absurd (mem_pair_imp_fst_mem_dom hz) hfx · intro hz
· intro hz exact absurd (mem_pair_imp_fst_mem_dom hz) hgx
exact absurd (mem_pair_imp_fst_mem_dom hz) hgx
/-- #### Exercise 3.15 /-- #### Exercise 3.15
@ -878,6 +878,174 @@ theorem exercise_3_15 {𝓐 : Set (Set.Relation α)}
have := hg' hg.right have := hg' hg.right
exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right exact single_valued_eq_unique (h𝓐 f hf.left) this hf.right
/-! #### Exercise 3.17
Show that the composition of two single-rooted sets is again single-rooted.
Conclude that the composition of two one-to-one functions is again one-to-one.
-/
theorem exercise_3_17_i {F G : Set.Relation α}
(hF : F.isSingleRooted) (hG : G.isSingleRooted)
: (F.comp G).isSingleRooted := by
intro v hv
have ⟨u₁, hu₁⟩ := ran_exists hv
have hu₁' := hu₁
unfold comp at hu₁'
simp only [Set.mem_setOf_eq] at hu₁'
have ⟨t₁, ht₁⟩ := hu₁'
unfold ExistsUnique
refine ⟨u₁, ⟨mem_pair_imp_fst_mem_dom hu₁, hu₁⟩, ?_⟩
intro u₂ hu₂
have hu₂' := hu₂
unfold comp at hu₂'
simp only [Set.mem_setOf_eq] at hu₂'
have ⟨_, ⟨t₂, ht₂⟩⟩ := hu₂'
have ht : t₁ = t₂ := single_rooted_eq_unique hF ht₁.right ht₂.right
rw [ht] at ht₁
exact single_rooted_eq_unique hG ht₂.left ht₁.left
theorem exercise_3_17_ii {F G : Set.Relation α}
(hF : F.isOneToOne) (hG : G.isOneToOne)
: (F.comp G).isOneToOne := And.intro
(single_valued_comp_is_single_valued hF.left hG.left)
(exercise_3_17_i hF.right hG.right)
/-! #### Exercise 3.18
Let `R` be the set
```
{⟨0, 1⟩, ⟨0, 2⟩, ⟨0, 3⟩, ⟨1, 2⟩, ⟨1, 3⟩, ⟨2, 3⟩}
```
Evaluate the following: `R ∘ R`, `R ↾ {1}`, `R⁻¹ ↾ {1}`, `R⟦{1}⟧`, and
`R⁻¹⟦{1}⟧`.
-/
section
variable {R : Set.Relation ℕ}
variable (hR : R = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)})
theorem exercise_3_18_i
: R.comp R = {(0, 2), (0, 3), (1, 3)} := by
rw [hR]
unfold comp
simp only [Set.mem_singleton_iff, Set.mem_insert_iff, or_self, Prod.mk.injEq]
ext x
have (a, b) := x
apply Iff.intro
· simp only [Set.mem_setOf_eq, Set.mem_singleton_iff, Set.mem_insert_iff]
intro ⟨t, ht₁, ht₂⟩
casesm* _ ∨ _
all_goals case _ hl hr => first
| {rw [hl.right] at hr; simp at hr}
| {rw [hl.left] at hr; simp at hr}
| {rw [hl.left, hr.right]; simp}
· simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
Prod.mk.injEq,
Set.mem_setOf_eq
]
intro h
casesm* _ ∨ _
· case _ h =>
refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
iterate 3 right
left
exact ⟨rfl, h.right⟩
· case _ h =>
refine ⟨1, Or.inl ⟨h.left, rfl⟩, ?_⟩
iterate 4 right
left
exact ⟨rfl, h.right⟩
· case _ h =>
refine ⟨2, ?_, ?_⟩
· iterate 3 right
left
exact ⟨h.left, rfl⟩
· iterate 5 right
exact ⟨rfl, h.right⟩
theorem exercise_3_18_ii
: R.restriction {1} = {(1, 2), (1, 3)} := by
rw [hR]
unfold restriction
ext p
have (a, b) := p
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
Set.mem_setOf_eq,
or_self
]
apply Iff.intro
· intro ⟨hp, ha⟩
rw [ha]
simp only [Prod.mk.injEq, true_and]
casesm* _ ∨ _
all_goals case _ h => first
| {rw [ha] at h; simp at h}
| {simp only [Prod.mk.injEq] at h; left; exact h.right}
| {simp only [Prod.mk.injEq] at h; right; exact h.right}
· intro h
apply Or.elim h
· intro hab
simp only [Prod.mk.injEq] at hab
refine ⟨?_, hab.left⟩
iterate 3 right
left
rw [hab.left, hab.right]
· intro hab
simp only [Prod.mk.injEq] at hab
refine ⟨?_, hab.left⟩
iterate 4 right
left
rw [hab.left, hab.right]
theorem exercise_3_18_iii
: R.inv.restriction {1} = {(1, 0)} := by
rw [hR]
unfold inv restriction
ext p
have (a, b) := p
simp only [
Set.mem_singleton_iff,
Set.mem_insert_iff,
or_self,
exists_eq_or_imp,
exists_eq_left,
Set.mem_setOf_eq,
Prod.mk.injEq
]
apply Iff.intro
· intro ⟨hb, ha⟩
casesm* _ ∨ _
all_goals case _ hr => first
| exact ⟨ha, hr.right.symm⟩
| rw [ha] at hr; simp at hr
· intro ⟨ha, hb⟩
rw [ha, hb]
simp
theorem exercise_3_18_iv
: R.image {1} = {2, 3} := by
rw [hR]
unfold image
ext y
simp
theorem exercise_3_18_v
: R.inv.image {1} = {0} := by
rw [hR]
unfold inv image
ext y
simp
end
end Relation end Relation
end Enderton.Set.Chapter_3 end Enderton.Set.Chapter_3