r
/
bookshelf
1
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Enderton. Ordering relations exercises.

finite-set-exercises
Joshua Potter 2023-07-14 14:06:53 -06:00
parent dd430d8379
commit 668c005c91
2 changed files with 190 additions and 8 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

194
Bookshelf/Enderton/Set.tex
View File

@ -5545,7 +5545,7 @@ State precisely the "analogous results" mentioned in Theorem 3Q.
\end{proof}
\subsection{\sorry{Exercise 3.43}}%
\subsection{\pending{Exercise 3.43}}%
\label{sub:exercise-3.43}
Assume that $R$ is a linear ordering on a set $A$.
@ -5553,25 +5553,124 @@ Show that $R^{-1}$ is also a linear ordering on $A$.
\begin{proof}
TODO
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}
Let $\pair{x, y}, \pair{y, z} \in R^{-1}$.
By definition of the \nameref{ref:inverse} of a set,
$\pair{y, x}$, $\pair{z, y} \in R$.
Since $R$ is transitive, it must be that $\pair{z, x} \in R$.
Then $\pair{x, z} \in R^{-1}$.
Thus $R^{-1}$ is transitive.
\paragraph{(ii)}%
\label{par:exercise-3.43-ii}
Let $x, y \in A$.
Since $R$ is trichotomous on $A$, it follows that exactly one of the
following conditions hold: $$xRy, \quad x = y, \quad yRx.$$
By definition of the \nameref{ref:inverse} of a set, the above possibilities
are equivalently expressed as
$$yR^{-1}x, \quad x = y, \quad xR^{-1}y.$$
Thus $R^{-1}$ is trichotomous.
\paragraph{Conclusion}%
Since $R^{-1}$ is transitive by \nameref{par:exercise-3.43-i} and
trichotomous by \nameref{par:exercise-3.43-ii}, it follows $R^{-1}$ is a
linear ordering on $A$.
\end{proof}
\subsection{\sorry{Exercise 3.44}}%
\subsection{\pending{Exercise 3.44}}%
\label{sub:exercise-3.44}
Assume that $<$ is a linear orderinng on a set $A$.
Assume that $<$ is a linear ordering on a set $A$.
Assume that $f \colon A \rightarrow A$ 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 $x < y$.
\begin{proof}
TODO
We show that (i) $f$ is one-to-one and (ii) whenever $f(x) < f(y)$, then
$x < y$.
\paragraph{(i)}%
Let $y \in \ran{f}$.
By definition of the \nameref{ref:range} of a set, there exists some
$x_1 \in A$ such that $f(x_1) = y$.
Suppose there exists some $x_2 \in A$ such that $f(x_2) = y$.
We prove $f$ is one-to-one by showing $x_1 = x_2$.
Because $<$ is a linear ordering on $A$, there exist three cases to
consider:
\subparagraph{Case 1}%
Assume $x_1 < x_2$.
By hypothesis, $f$ is monotonic.
Thus $f(x_1) < f(x_2)$.
But $<$ is a trichotomous relation meaning it is not possible for
\textit{both} $f(x) = f(y)$ and $f(x) < f(y)$.
Thus our original assumption must be wrong.
\subparagraph{Case 2}%
Assume $x_1 = x_2$.
Then we are immediately finished.
\subparagraph{Case 3}%
Assume $x_1 > x_2$.
By hypothesis, $f$ is monotonic.
Thus $f(x_1) > f(x_2)$.
But $<$ is a trichotomous relation meaning it is not possible for
\textit{both} $f(x) = f(y)$ and $f(x) > f(y)$.
Thus our original assumption must be wrong.
\subparagraph{Conclusion}%
Since the above cases are exhaustive, the only possibility is $x_1 = x_2$.
Thus $f$ is one-to-one.
\paragraph{(ii)}%
Suppose $f(x) < f(y)$.
There are three cases to consider:
\subparagraph{Case 1}%
Assume $x < y$.
Then we are immediately finished.
\subparagraph{Case 2}%
Assume $x = y$.
Then $f(x) = f(y)$.
But $<$ is a trichotomous relation meaning it is not possible for
\textit{both} $f(x) = f(y)$ and $f(x) < f(y)$.
Thus our original assumption must be wrong.
\subparagraph{Case 3}%
Assume $x > y$.
By hypothesis, $f$ is monotonic.
Thus $f(x) > f(y)$.
But $<$ is a trichotomous relation meaning it is not possible for
\textit{both} $f(x) < f(y)$ and $f(x) > f(y)$.
Thus our original assumption must be wrong.
\subparagraph{Conclusion}%
Since the above cases are exhaustive, the only possibility is $x < y$.
\end{proof}
\subsection{\sorry{Exercise 3.45}}%
\subsection{\pending{Exercise 3.45}}%
\label{sub:exercise-3.45}
Assume that $<_A$ and $<_B$ are linear orderings on $A$ and $B$, respectively.
@ -5584,7 +5683,88 @@ Show that $<_L$ is a linear ordering on $A \times B$.
\begin{proof}
TODO
We show that $<_L$ is (i) \nameref{ref:transitive} and (ii)
\nameref{ref:trichotomous} on $A \times B$.
\paragraph{(i)}%
Let $\pair{a_1, b_1} <_L \pair{a_2, b_2}$ and
$\pair{a_2, b_2} <_L \pair{a_3, b_3}$.
Then either $a_1 <_A a_2$ or $a_1 = a_2 \land b_1 <_B b_2$.
Likewise, either $a_2 <_A a_3$ or $a_2 = a_3 \land b_2 <_B b_3$.
We consider each combination of cases in turn:
\subparagraph{Case 1}%
Suppose $a_1 <_A a_2$ and $a_2 <_A a_3$.
Since $<_A$ is a linear ordering, it follows $<_A$ is transitive.
Thus $a_1 <_A a_3$.
Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$.
\subparagraph{Case 2}%
Suppose $a_1 <_A a_2$, $a_2 = a_3$, and $b_2 <_B b_3$.
Then $a_1 < a_3$.
Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$.
\subparagraph{Case 3}%
Suppose $a_1 = a_2$, $b_1 <_B b_2$, and $a_2 <_A a_3$.
Then $a_1 <_A a_3$.
Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$.
\subparagraph{Case 4}%
Suppose $a_1 = a_2$, $b_1 <_B b_2$, $a_2 = a_3$, and $b_2 <_B b_3$.
Then $a_1 = a_3$.
Since $<_B$ is a linear ordering, it follows $<_B$ is transitive.
Thus $b_1 <_B b_3$.
Therefore $\pair{a_1, b_2} <_L \pair{a_3, b_3}$.
\subparagraph{Conclusion}%
These four cases are exhaustive and each conclude that $<_L$ is
transitive.
\paragraph{(ii)}%
Let $\pair{a_1, b_1}, \pair{a_2, b_2} \in A \times B$.
Because $<_A$ and $<_B$ are linear orderings on $A$ and $B$ respectively,
it follows $<_A$ and $<_B$ are both trichotomous on their respective sets.
Thus exactly one of $$a_1 <_A a_2, \quad a_1 = a_2, \quad a_2 <_A a_1$$
and $$b_1 <_B b_2, \quad b_1 = b_2, \quad b_2 <_B b_1$$ holds.
There are three cases we examine:
\subparagraph{Case 1}%
\label{spar:exercise-3.45-ii-case-1}
Suppose $a_1 <_A a_2$.
Then $\pair{a_1, b_1} <_L \pair{a_2, b_2}$.
This is trivially the only possible relationship between the ordered
pairs.
\subparagraph{Case 2}%
Suppose $a_1 = a_2$.
If $b_1 <_B b_2$, then $\pair{a_1, b_1} <_L \pair{a_2, b_2}$ is the only
possibility.
If $b_1 = b_2$, then $\pair{a_1, b_1} = \pair{a_2, b_2}$ is the only
possibility.
If $b_2 <_B b_1$, then $\pair{a_2, b_2} <_L \pair{a_1, b_1}$ is the only
possibility.
\subparagraph{Case 3}%
Suppose $a_2 <_A a_1$.
This case is analagous to \eqref{spar:exercise-3.45-ii-case-1}.
\subparagraph{Conclusion}%
In each of the above cases, we are always left with exactly one of
$$\pair{a_1, b_1} <_L \pair{a_2, b_2}, \quad
\pair{a_1, b_1} = \pair{a_2, b_2}, \quad
\pair{a_2, b_2} <_L \pair{a_1, b_1}.$$
Thus $<_L$ is trichotomous.
\end{proof}

4
Bookshelf/Enderton/Set/Chapter_3.lean
View File

@ -3,6 +3,8 @@ import Bookshelf.Enderton.Set.OrderedPair
import Bookshelf.Enderton.Set.Relation
import Common.Logic.Basic
import Mathlib.Data.Real.Basic
import Mathlib.Data.Rel
import Mathlib.Order.RelClasses
import Mathlib.Tactic.CasesM
/-! # Enderton.Set.Chapter_3
@ -2109,7 +2111,7 @@ theorem exercise_3_39 {P : Set (Set α)} {R Rₚ : Set.Relation α} {A : Set α}
refine ⟨neighborhood R x, ?_, ⟨hx, hy⟩⟩
rw [hP]
exact ⟨x, hxA, rfl⟩
lemma test {x y z : ℝ} (h : x + y = z) : (x = z - y) := by apply?
/-- #### Exercise 3.41 (a)
Let `ℝ` be the set of real numbers and define the realtion `Q` on `ℝ × ℝ` by