diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex index 4ce1ff9..2596d9c 100644 --- a/Bookshelf/Apostol.tex +++ b/Bookshelf/Apostol.tex @@ -185,8 +185,10 @@ That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$ such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$ Step functions are sometimes called \textbf{piecewise constant functions}. -\note{At each of the endpoints $x_{k-1}$ and $x_k$ the function must - have some well-defined value, but this need not be the same as $s_k$.} +\begin{note} + At each of the endpoints $x_{k-1}$ and $x_k$ the function must have some + well-defined value, but this need not be the same as $s_k$. +\end{note} \begin{definition} @@ -478,7 +480,9 @@ Let $h$ be a given positive number and let $S$ be a set of real numbers. Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set $$C = \{a + b : a \in A, b \in B\}.$$ -\note{This is known as the "Additive Property."} +\begin{note} + This is known as the "Additive Property." +\end{note} \subsection{\verified{Theorem I.33a}}% \hyperlabel{sub:theorem-i.33a} @@ -1609,8 +1613,10 @@ If $a$ and $b$ are positive integers with no common factor, we have the formula $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$ When $b = 1$, the sum on the left is understood to be $0$. -\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the - assumption $b > 1$.} +\begin{note} + When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the + assumption $b > 1$. +\end{note} \subsubsection{\pending{Exercise 1.11.7a}}% \hyperlabel{ssub:exercise-1.11.7a} @@ -2394,8 +2400,10 @@ Which of the following properties would remain valid in this new theory? $\int_a^b s + \int_b^c s = \int_a^c s$. -\note{This property mirrors - \nameref{sub:step-additivity-with-respect-interval-integration}.} +\begin{note} + This property mirrors + \nameref{sub:step-additivity-with-respect-interval-integration}. +\end{note} \begin{proof} @@ -2430,7 +2438,9 @@ $\int_a^b s + \int_b^c s = \int_a^c s$. $\int_a^b (s + t) = \int_a^b s + \int_a^b t$. -\note{This property mirrors the \nameref{sub:step-additive-property}.} +\begin{note} + This property mirrors the \nameref{sub:step-additive-property}. +\end{note} \begin{proof} @@ -2477,7 +2487,9 @@ $\int_a^b (s + t) = \int_a^b s + \int_a^b t$. $\int_a^b c \cdot s = c \int_a^b s$. -\note{This property mirrors the \nameref{sub:step-homogeneous-property}.} +\begin{note} + This property mirrors the \nameref{sub:step-homogeneous-property}. +\end{note} \begin{proof} @@ -2509,7 +2521,9 @@ $\int_a^b c \cdot s = c \int_a^b s$. $\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$. -\note{This property mirrors \nameref{sub:step-invariance-under-translation}.} +\begin{note} + This property mirrors \nameref{sub:step-invariance-under-translation}. +\end{note} \begin{proof} @@ -2547,7 +2561,9 @@ $\int_{a+c}^{b+c} s(x) \mathop{dx} = \int_a^b s(x + c) \mathop{dx}$. If $s(x) < t(x)$ for each $x$ in $[a, b]$, then $\int_a^b s < \int_a^b t$. -\note{This property mirrors the \nameref{sub:step-comparison-theorem}.} +\begin{note} + This property mirrors the \nameref{sub:step-comparison-theorem}. +\end{note} \begin{proof} diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex index 0d5594d..3375056 100644 --- a/Bookshelf/Enderton/Set.tex +++ b/Bookshelf/Enderton/Set.tex @@ -255,9 +255,26 @@ A \textbf{linear ordering} on $A$ (also called a \textbf{total ordering} on $A$) \item $R$ is \nameref{ref:trichotomous}. \end{enumerate} +\begin{note} + This definition does not agree with how Lean defines a linear order. + + \vspace{6pt} + Trichotomy is equivalent to asymmetry and connectivity and asymmetry is + equivalent to antisymmetry and irreflexivity. Thus a linear order, as defined + by Enderton, is a binary relation with the following four properties: + + \vspace{6pt} + \begin{enumerate}[(i)] + \item Irreflexivity + \item Antisymmetry + \item Connectivity (i.e. totality) + \item Transitivity + \end{enumerate} +\end{note} + \begin{definition} - \lean*{Mathlib/Init/Algebra/Classes}{IsLinearOrder} + \lean*{Mathlib/Init/Algebra/Classes}{IsStrictTotalOrder} \end{definition} @@ -811,7 +828,9 @@ List all the members of $V_4$. There is no set to which every set belongs. - \note{This was revisited after reading Enderton's proof prior.} + \begin{note} + This was revisited after reading Enderton's proof prior. + \end{note} \end{theorem} @@ -2670,8 +2689,10 @@ If not, then under what conditions does equality hold? \lean*{Mathlib/SetTheory/ZFC/Basic}{Set.prod} - \note{The above Lean proof is a definition (i.e. an axiom). It does not prove - such a set's existence from first principles.} + \begin{note} + The above Lean proof is a definition (i.e. an axiom). It does not prove + such a set's existence from first principles. + \end{note} Define $C = A \cup B$. Then for all $x \in A$ and for all $y \in B$, $x$ and $y$ are both in $C$. @@ -3556,7 +3577,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \section{Ordering Relations}% \hyperlabel{sec:ordering-relations} -\subsection{\pending{Theorem 3R}}% +\subsection{\verified{Theorem 3R}}% \hyperlabel{sub:theorem-3r} \begin{theorem}[3R] @@ -3571,6 +3592,9 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one. \begin{proof} + \lean{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.theorem\_3r} + Suppose $R$ is a \nameref{ref:linear-ordering} on $A$. \paragraph{(i)}% @@ -5545,20 +5569,23 @@ State precisely the "analogous results" mentioned in Theorem 3Q. \end{proof} -\subsection{\pending{Exercise 3.43}}% -\label{sub:exercise-3.43} +\subsection{\verified{Exercise 3.43}}% +\hyperlabel{sub:exercise-3.43} Assume that $R$ is a linear ordering on a set $A$. Show that $R^{-1}$ is also a linear ordering on $A$. \begin{proof} + \lean{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.exercise\_3\_43} + Assume that $R$ is a \nameref{ref:linear-ordering} on a set $A$. Then $R$ is \nameref{ref:transitive} and \nameref{ref:trichotomous}. We show that (i) $R^{-1}$ is transitive and (ii) $R^{-1}$ is trichotomous. \paragraph{(i)}% - \label{par:exercise-3.43-i} + \hyperlabel{par:exercise-3.43-i} Let $\pair{x, y}, \pair{y, z} \in R^{-1}$. By definition of the \nameref{ref:inverse} of a set, @@ -5568,7 +5595,7 @@ Show that $R^{-1}$ is also a linear ordering on $A$. Thus $R^{-1}$ is transitive. \paragraph{(ii)}% - \label{par:exercise-3.43-ii} + \hyperlabel{par:exercise-3.43-ii} Let $x, y \in A$. Since $R$ is trichotomous on $A$, it follows that exactly one of the @@ -5586,8 +5613,8 @@ Show that $R^{-1}$ is also a linear ordering on $A$. \end{proof} -\subsection{\pending{Exercise 3.44}}% -\label{sub:exercise-3.44} +\subsection{\verified{Exercise 3.44}}% +\hyperlabel{sub:exercise-3.44} Assume that $<$ is a linear ordering on a set $A$. Assume that $f \colon A \rightarrow A$ and that $f$ has the property that @@ -5596,6 +5623,14 @@ Show that $f$ is one-to-one and that whenever $f(x) < f(y)$, then $x < y$. \begin{proof} + \statementpadding + + \lean*{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.exercise\_3\_44\_i} + + \lean{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.exercise\_3\_44\_ii} + We show that (i) $f$ is one-to-one and (ii) whenever $f(x) < f(y)$, then $x < y$. @@ -5670,8 +5705,8 @@ Show that $f$ is one-to-one and that whenever $f(x) < f(y)$, then $x < y$. \end{proof} -\subsection{\pending{Exercise 3.45}}% -\label{sub:exercise-3.45} +\subsection{\verified{Exercise 3.45}}% +\hyperlabel{sub:exercise-3.45} Assume that $<_A$ and $<_B$ are linear orderings on $A$ and $B$, respectively. Define the binary relation $<_L$ on the Cartesian product $A \times B$ by: @@ -5683,6 +5718,9 @@ Show that $<_L$ is a linear ordering on $A \times B$. \begin{proof} + \lean{Bookshelf/Enderton/Set/Chapter\_3} + {Enderton.Set.Chapter\_3.exercise\_3\_45} + We show that $<_L$ is (i) \nameref{ref:transitive} and (ii) \nameref{ref:trichotomous} on $A \times B$. @@ -5736,7 +5774,7 @@ Show that $<_L$ is a linear ordering on $A \times B$. There are three cases we examine: \subparagraph{Case 1}% - \label{spar:exercise-3.45-ii-case-1} + \hyperlabel{spar:exercise-3.45-ii-case-1} Suppose $a_1 <_A a_2$. Then $\pair{a_1, b_1} <_L \pair{a_2, b_2}$. diff --git a/Bookshelf/Enderton/Set/Chapter_3.lean b/Bookshelf/Enderton/Set/Chapter_3.lean index f6babb7..f6ac7bf 100644 --- a/Bookshelf/Enderton/Set/Chapter_3.lean +++ b/Bookshelf/Enderton/Set/Chapter_3.lean @@ -4,6 +4,7 @@ import Bookshelf.Enderton.Set.Relation import Common.Logic.Basic import Mathlib.Data.Real.Basic import Mathlib.Data.Rel +import Mathlib.Init.Algebra.Classes import Mathlib.Order.RelClasses import Mathlib.Tactic.CasesM @@ -2167,4 +2168,174 @@ theorem exercise_3_41_a {Q : Set.Relation (ℝ × ℝ)} end Relation +/-- #### Theorem 3R + +Let `R` be a linear ordering on `A`. + +(i) There is no `x` for which `xRx`. +(ii) For distinct `x` and `y` in `A`, either `xRy` or `yRx`. +-/ +theorem theorem_3r {R : Rel α α} (hR : IsStrictTotalOrder α R) + : (∀ x : α, ¬ R x x) ∧ (∀ x y : α, x ≠ y → R x y ∨ R y x) := by + apply And.intro + · exact hR.irrefl + · intro x y h + apply Or.elim (hR.trichotomous x y) + · intro h₁ + left + exact h₁ + · intro h₁ + apply Or.elim h₁ + · intro h₂ + exact absurd h₂ h + · intro h₂ + right + exact h₂ + +/-- #### Exercise 3.43 + +Assume that `R` is a linear ordering on a set `A`. Show that `R⁻¹` is also a +linear ordering on `A`. +-/ +theorem exercise_3_43 {R : Rel α α} (hR : IsStrictTotalOrder α R) + : IsStrictTotalOrder α R.inv := by + refine { trichotomous := ?_, irrefl := ?_, trans := ?_ } + · intro a b + unfold Rel.inv flip + apply Or.elim (hR.trichotomous a b) + · intro h; right; right; exact h + · intro h + apply Or.elim h + · intro h; right; left; exact h + · intro h; left; exact h + · intro x h + unfold Rel.inv flip at h + exact absurd h (hR.irrefl x) + · intro a b c hab hac + unfold Rel.inv flip at * + exact hR.trans c b a hac hab + +/-! #### Exercise 3.44 + +Assume that `<` is a linear ordering on a set `A`. Assume that `f : A → A` and +that `f` has the property that whenever `x < y`, then `f(x) < f(y)`. Show that +`f` is one-to-one and that whenever `f(x) < f(y)`, then `x < y`. +-/ + +theorem exercise_3_44_i {R : Rel α α} (hR : IsStrictTotalOrder α R) + (f : α → α) (hf : ∀ x y, R x y → R (f x) (f y)) + : Function.Injective f := by + unfold Function.Injective + intro x₁ x₂ hx + apply Or.elim (hR.trichotomous x₁ x₂) + · -- `x₁ < x₂` + intro hx₁ + have nh := hf x₁ x₂ hx₁ + rw [hx] at nh + exact absurd nh (hR.irrefl (f x₂)) + · intro hx₁ + apply Or.elim hx₁ + · simp -- `x₁ = x₂` + · -- `x₁ > x₂` + intro hx₂ + have nh := hf x₂ x₁ hx₂ + rw [← hx] at nh + exact absurd nh (hR.irrefl (f x₁)) + +theorem exercise_3_44_ii {R : Rel α α} (hR : IsStrictTotalOrder α R) + (f : α → α) (hf : ∀ x y, R x y → R (f x) (f y)) + : R (f x) (f y) → R x y := by + intro h + apply Or.elim (hR.trichotomous x y) + · simp -- `x < y` + · intro h₁ + apply Or.elim h₁ + · -- `x = y` + intro h₂ + rw [h₂] at h + exact absurd h (hR.irrefl (f y)) + · -- `x > y` + intro h₂ + have := hR.trans (f x) (f y) (f x) h (hf y x h₂) + exact absurd this (hR.irrefl (f x)) + +/-- #### Exercise 3.45 + +Assume that `<_A` and `<_B` are linear orderings on `A` and `B`, respectively. +Define the binary relation `<_L` on the Cartesian product `A × B` by: +``` +⟨a₁, b₁⟩ <_L ⟨a₂, b₂⟩ iff either a₁ <_A a₂ or (a₁ = a₂ ∧ b₁ <_B b₂). +``` +Show that `<_L` is a linear ordering on `A × B`. (The relation `<_L` is called +*lexicographic* ordering, being the ordering used in making dictionaries.) +-/ +theorem exercise_3_45 {A : Rel α α} {B : Rel β β} {R : Rel (α × β) (α × β)} + (hA : IsStrictTotalOrder α A) (hB : IsStrictTotalOrder β B) + (hR : ∀ a₁ b₁ a₂ b₂, R (a₁, b₁) (a₂, b₂) ↔ A a₁ a₂ ∨ (a₁ = a₂ ∧ B b₁ b₂)) + : IsStrictTotalOrder (α × β) R := by + refine { trichotomous := ?_, irrefl := ?_, trans := ?_ } + · intro (a₁, b₁) (a₂, b₂) + apply Or.elim (hA.trichotomous a₁ a₂) + · -- `a₁ <_A a₂` + intro ha + left + exact (hR a₁ b₁ a₂ b₂).mpr (Or.inl ha) + · intro ha + apply Or.elim ha + · -- `a₁ = a₂` + intro ha₁ + apply Or.elim (hB.trichotomous b₁ b₂) + · -- `b₁ <_B b₂` + intro hb + left + exact (hR a₁ b₁ a₂ b₂).mpr (Or.inr ⟨ha₁, hb⟩) + · intro hb + apply Or.elim hb + · -- `b₁ = b₂` + intro hb₁ + right; left + rw [ha₁, hb₁] + · -- `b₂ <_B b₁` + intro hb₁ + right; right + exact (hR a₂ b₂ a₁ b₁).mpr (Or.inr ⟨ha₁.symm, hb₁⟩) + · -- `a₂ <_A a₁` + intro ha₁ + right; right + exact (hR a₂ b₂ a₁ b₁).mpr (Or.inl ha₁) + · intro (a, b) h + have := (hR a b a b).mp h + apply Or.elim this + · intro ha₁ + exact absurd ha₁ (hA.irrefl a) + · intro ⟨_, hb₁⟩ + exact absurd hb₁ (hB.irrefl b) + · intro (a₁, b₁) (a₂, b₂) (a₃, b₃) h₁ h₂ + have h₁' := (hR a₁ b₁ a₂ b₂).mp h₁ + have h₂' := (hR a₂ b₂ a₃ b₃).mp h₂ + apply Or.elim h₁' + · -- `a₁ <_A a₂` + intro ha₁ + apply Or.elim h₂' + · -- `a₂ <_A a₃` + intro ha₂ + have := hA.trans a₁ a₂ a₃ ha₁ ha₂ + exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl this) + · -- `a₂ = a₃ ∧ b₂ <_B b₃` + intro ha₂ + rw [ha₂.left] at ha₁ + exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl ha₁) + · -- `a₁ = a₂ ∧ b₁ <_B b₂` + intro ha₁ + apply Or.elim h₂' + · -- `a₂ <_A a₃` + intro ha₂ + rw [← ha₁.left] at ha₂ + exact (hR a₁ b₁ a₃ b₃).mpr (Or.inl ha₂) + · -- `a₂ = a₃ ∧ b₂ <_B b₃` + intro ⟨ha₂, hb₂⟩ + rw [← ha₁.left] at ha₂ + have := hB.trans b₁ b₂ b₃ ha₁.right hb₂ + exact (hR a₁ b₁ a₃ b₃).mpr (Or.inr ⟨ha₂, this⟩) + end Enderton.Set.Chapter_3 \ No newline at end of file diff --git a/preamble.tex b/preamble.tex index 7443eb7..3b4eec1 100644 --- a/preamble.tex +++ b/preamble.tex @@ -83,8 +83,8 @@ \end{minipage}} \end{center}} -\newcommand{\note}[1]{\@admonition{Note:}{#1}} -\newcommand{\todo}[1]{\@admonition{TODO:}{#1}} +\NewEnviron{note}{\@admonition{NOTE:}{\BODY}} +\NewEnviron{todo}{\@admonition{TODO:}{\BODY}} % ======================================== % Statements