Enderton. Finishe exercise set 5. Prep for exercise set 6.

finite-set-exercises
Joshua Potter 2023-06-15 15:31:58 -06:00
parent f5d1fc546a
commit 2abd4b6b25
3 changed files with 193 additions and 47 deletions

View File

@ -2519,66 +2519,58 @@ Show that there is no set to which every ordered pair belongs.
\end{proof}
\subsection{\unverified{Exercise 5.5a}}%
\subsection{\verified{Exercise 5.5a}}%
\label{sub:exercise-5.5a}
Assume that $A$ and $B$ are given sets, and show that there exists a set $C$
such that for any $y$,
$$y \in C \iff y = \{x\} \times B \text{ for some } x \text{ in } A.$$
\begin{equation}
\label{sub:exercise-5.5a-eq1}
y \in C \iff y = \{x\} \times B \text{ for some } x \text{ in } A.
\end{equation}
In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set.
\begin{proof}
Let $A$ and $B$ be arbitrary sets.
Also let $x \in A$.
By definition of the \nameref{sub:cartesian-product},
\begin{equation}
\label{sub:exercise-5.5a-eq1}
\{x\} \times B = \{ \left< x, b \right> \mid b \in B \}.
\end{equation}
If $B = \emptyset$ then $\{x\} \times B$ trivially evaluates to the empty set,
which is a set by virtue of the \nameref{ref:empty-set-axiom}.
Therefore we continue under the assumption $B \neq \emptyset$.
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.exercise\_5\_5a}
We prove that (i)
$\{x\} \times B \subseteq \powerset{\powerset{(\{x\} \cup B)}}$ and then
(ii) that $\{x\} \times B$ is a set.
Let $a \in A$.
By the \nameref{ref:pairing-axiom}, $\{a\}$ is a set.
By \nameref{sub:cartesian-product}, $\{a\} \times B$ is a set.
Again by the \nameref{ref:pairing-axiom}, $\{\{a\} \times B\}$ is a set.
\paragraph{(i)}%
\label{par:exercise-5.5a-i}
Next, by another application of \nameref{sub:cartesian-product}, $A \times B$
is a set.
By the \nameref{ref:power-set-axiom}, $\powerset{(A \times B)}$ is a set.
Thus, by the \nameref{ref:subset-axioms}, the following is also a set:
$$C = \{ y \in \powerset{(A \times B)} \mid
\exists a \in A, \forall x, \left[ x \in y \iff
\exists b \in B, x = \left< a, b \right> \right] \}.$$
We now show that $C$ satisfies \eqref{sub:exercise-5.5a-eq1}.
Let $t \in \{x\} \times B$.
By \eqref{sub:exercise-5.5a-eq1} and definition of an
\nameref{ref:ordered-pair}, there exists a $b \in B$ such that
$$t = \left< x, b \right> = \{\{x\}, \{x, b\}\}.$$
It trivially holds that
$$\{x\} \subseteq \{x\} \cup B \quad\text{and}\quad
\{x, b\} \subseteq \{x\} \cup B.$$
Therefore, by definition of the \nameref{ref:power-set},
$$\{x\} \in \powerset{(\{x\} \cup B)} \quad\text{and}\quad
\{x, b\} \in \powerset{(\{x\} \cup B)}.$$
But then $\{\{x\}, \{x, b\}\} \subseteq \powerset{(\{x\} \cup B)}$.
Another application of the definition of the \nameref{ref:power-set} implies
that $$\{\{x\}, \{x, b\}\} \in \powerset{\powerset{(\{x\} \cup B)}}.$$
Since this holds for all sets $t$,
$\{x\} \times B \subseteq \powerset{\powerset{(\{x\} \cup B)}}$.
\paragraph{($\Rightarrow$)}%
\paragraph{(ii)}%
\label{par:exercise-5.5a-ii}
Suppose $y \in C$.
Then there exists some $a \in A$ such that
$$\forall x, \left[ x \in y \iff
\exists b \in B, x = \left< a, b \right> \right].$$
By the \nameref{ref:extensionality-axiom},
\begin{align*}
y
& = \{ \left< a, b \right> \mid b \in B \} \\
& = \{ \left< x, b \right> \mid x \in \{a\} \land b \in B \} \\
& = \{ \{a\} \times B \}.
\end{align*}
By the \nameref{ref:subset-axioms}, there exists a set $C$ such that for any
set $y$, $$y \in C \iff
y \in \powerset{\powerset{(\{x\} \cup B)}} \land
(\exists b \in B, y = \left< x, b \right>).$$
The above equation and \eqref{sub:exercise-5.5a-eq1} implies $C$ contains
only ordered pairs of the desired form.
By \nameref{par:exercise-5.5a-i}, $C$ contains them all.
\paragraph{($\Leftarrow$)}%
\paragraph{Conclusion}%
Since $x$ was an arbitrarily chosen member of $A$,
\nameref{par:exercise-5.5a-ii} immediately implies that
$\{\{x\} \times B \mid x \in A\}$ is indeed a set.
Suppose $y = \{a\} \times B$ for some $a \in A$.
By \nameref{sub:cartesian-product}, $x \in \{a\} \times B$ if and only if
$\exists b \in B$ such that $x = \left< a, b \right>$.
But then $x \in y$ if and only if $\exists b \in B$ such that
$x = \left< a, b \right>$.
This immediately proves $y \in C$.
\end{proof}
@ -2633,4 +2625,87 @@ With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$.
\end{proof}
\section{Relations}%
\label{sec:relations}
\subsection{\verified{Theorem 3D}}%
\label{sub:theorem-3d}
\begin{theorem}[3D]
If $\left< x, y \right> \in A$, then $x$ and $y$ belong to $\bigcup\bigcup A$.
\end{theorem}
\begin{proof}
\lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3d}
Let $A$ be a set and $\left< x, y \right> \in A$.
By definition of an \nameref{ref:ordered-pair},
$$\left< x, y \right> = \{\{x\}, \{x, y\}\}.$$
By \nameref{sub:exercise-3.3}, $\{\{x\}, \{x, y\}\} \subseteq \bigcup A$.
Then $\{x, y\} \in \bigcup A$.
Another application of \nameref{sub:exercise-3.3} implies
$\{x, y\} \subseteq \bigcup\bigcup A$.
Therefore $x, y \in \bigcup\bigcup A$.
\end{proof}
\section{Exercises 6}%
\label{sec:exercises-6}
\subsection{\unverified{Exercise 6.6}}%
\label{sub:exercise-6.6}
Show that a set $A$ is a relation iff
$A \subseteq \textop{dom} A \times \textop{ran} A$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 6.7}}%
\label{sub:exercise-6.7}
Show that if $R$ is a relation, then $\textop{fld} R = \bigcup\bigcup R$.
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 6.8}}%
\label{sub:exercise-6.8}
Show that for any set $\mathscr{A}$:
\begin{align*}
\textop{dom} \bigcup{\mathscr{A}}
& = \bigcup\; \{ \textop{dom} R \mid R \in \mathscr{A} \},
\textop{ran} \bigcup{\mathscr{A}}
& = \bigcup\; \{ \textop{ran} R \mid R \in \mathscr{A} \}.
\end{align*}
\begin{proof}
TODO
\end{proof}
\subsection{\unverified{Exercise 6.9}}%
\label{sub:exercise-6.9}
Discuss the result of replacing the union operation by the intersection
operation in the preceding problem.
\begin{proof}
TODO
\end{proof}
\end{document}

View File

@ -1,5 +1,6 @@
import Mathlib.Data.Set.Basic
import Bookshelf.Enderton.Set.Chapter_2
import Common.Logic.Basic
import Common.Set.Basic
@ -221,7 +222,61 @@ In other words, show that `{{x} × B | x ∈ A}` is a set.
theorem exercise_5_5a {A : Set α} {B : Set β}
: ∃ C : Set (Set (α × β)),
y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
sorry
let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
refine ⟨C, ?_⟩
apply Iff.intro
· intro hC
simp only [Set.mem_setOf_eq] at hC
have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
refine ⟨a, ⟨ha, ?_⟩⟩
ext x
apply Iff.intro
· intro hxy
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
rw [Prod.ext_iff] at hx
simp only at hx
rw [← hx.right] at hb
exact ⟨hx.left, hb⟩
· intro hx
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
have hxab : x = (a, x.snd) := by
ext <;> simp
exact hx.left
rwa [← hxab] at this
· intro ⟨x, ⟨hx, hy⟩⟩
show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
apply And.intro
· simp only [Set.mem_powerset_iff]
rw [hy]
unfold Set.prod
simp only [
Set.mem_singleton_iff,
Set.setOf_subset_setOf,
and_imp,
Prod.forall
]
intro a b ha hb
exact ⟨by rw [ha]; exact hx, hb⟩
· refine ⟨x, ⟨hx, ?_⟩⟩
intro p
apply Iff.intro
· intro hab
rw [hy] at hab
unfold Set.prod at hab
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
· intro ⟨b, ⟨hb, hab⟩⟩
rw [hy]
unfold Set.prod
simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
rw [Prod.ext_iff] at hab
simp only at hab
rw [hab.right]
exact ⟨hab.left, hb⟩
/-- ### Exercise 5.5b
@ -258,4 +313,19 @@ theorem exercise_5_5b {A : Set α} (B : Set β)
rw [← ha] at h
exact ⟨h, hb⟩
/-- ### Theorem 3D
If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to ` A`.
-/
theorem theorem_3d {A : Set (Set (Set (αα)))} (h : OrderedPair x y ∈ A)
: Sum.inl x ∈ ⋃₀ (⋃₀ A) ∧ Sum.inr y ∈ ⋃₀ (⋃₀ A) := by
have hp : OrderedPair x y ⊆ ⋃₀ A := Chapter_2.exercise_3_3 (OrderedPair x y) h
have hp' : ∀ t, t ∈ {{Sum.inl x}, {Sum.inl x, Sum.inr y}} → t ∈ ⋃₀ A := hp
have hq := hp' {Sum.inl x, Sum.inr y} (by simp)
have hq' := Chapter_2.exercise_3_3 {Sum.inl x, Sum.inr y} hq
have : ∀ t, t ∈ {Sum.inl x, Sum.inr y} → t ∈ ⋃₀ (⋃₀ A) := hq'
exact ⟨this (Sum.inl x) (by simp), this (Sum.inr y) (by simp)⟩
end Enderton.Set.Chapter_3

View File

@ -134,6 +134,7 @@
\newcommand{\ioc}[2]{\left(#1, #2\right]}
\newcommand{\ioo}[2]{\left(#1, #2\right)}
\newcommand{\powerset}[1]{\mathscr{P}#1}
\newcommand{\textop}[1]{\mathop{\text{#1}}}
\newcommand{\ubar}[1]{\text{\b{$#1$}}}
\let\oldemptyset\emptyset