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