Enderton. Finishe exercise set 5. Prep for exercise set 6.
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@ -2519,66 +2519,58 @@ Show that there is no set to which every ordered pair belongs.
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\end{proof}
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\subsection{\unverified{Exercise 5.5a}}%
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\subsection{\verified{Exercise 5.5a}}%
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\label{sub:exercise-5.5a}
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Assume that $A$ and $B$ are given sets, and show that there exists a set $C$
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such that for any $y$,
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$$y \in C \iff y = \{x\} \times B \text{ for some } x \text{ in } A.$$
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\begin{equation}
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\label{sub:exercise-5.5a-eq1}
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y \in C \iff y = \{x\} \times B \text{ for some } x \text{ in } A.
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\end{equation}
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In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set.
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\begin{proof}
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Let $A$ and $B$ be arbitrary sets.
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Also let $x \in A$.
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By definition of the \nameref{sub:cartesian-product},
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\begin{equation}
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\label{sub:exercise-5.5a-eq1}
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\{x\} \times B = \{ \left< x, b \right> \mid b \in B \}.
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\end{equation}
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If $B = \emptyset$ then $\{x\} \times B$ trivially evaluates to the empty set,
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which is a set by virtue of the \nameref{ref:empty-set-axiom}.
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Therefore we continue under the assumption $B \neq \emptyset$.
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_5\_5a}
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We prove that (i)
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$\{x\} \times B \subseteq \powerset{\powerset{(\{x\} \cup B)}}$ and then
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(ii) that $\{x\} \times B$ is a set.
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Let $a \in A$.
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By the \nameref{ref:pairing-axiom}, $\{a\}$ is a set.
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By \nameref{sub:cartesian-product}, $\{a\} \times B$ is a set.
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Again by the \nameref{ref:pairing-axiom}, $\{\{a\} \times B\}$ is a set.
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\paragraph{(i)}%
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\label{par:exercise-5.5a-i}
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Next, by another application of \nameref{sub:cartesian-product}, $A \times B$
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is a set.
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By the \nameref{ref:power-set-axiom}, $\powerset{(A \times B)}$ is a set.
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Thus, by the \nameref{ref:subset-axioms}, the following is also a set:
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$$C = \{ y \in \powerset{(A \times B)} \mid
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\exists a \in A, \forall x, \left[ x \in y \iff
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\exists b \in B, x = \left< a, b \right> \right] \}.$$
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We now show that $C$ satisfies \eqref{sub:exercise-5.5a-eq1}.
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Let $t \in \{x\} \times B$.
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By \eqref{sub:exercise-5.5a-eq1} and definition of an
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\nameref{ref:ordered-pair}, there exists a $b \in B$ such that
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$$t = \left< x, b \right> = \{\{x\}, \{x, b\}\}.$$
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It trivially holds that
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$$\{x\} \subseteq \{x\} \cup B \quad\text{and}\quad
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\{x, b\} \subseteq \{x\} \cup B.$$
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Therefore, by definition of the \nameref{ref:power-set},
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$$\{x\} \in \powerset{(\{x\} \cup B)} \quad\text{and}\quad
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\{x, b\} \in \powerset{(\{x\} \cup B)}.$$
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But then $\{\{x\}, \{x, b\}\} \subseteq \powerset{(\{x\} \cup B)}$.
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Another application of the definition of the \nameref{ref:power-set} implies
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that $$\{\{x\}, \{x, b\}\} \in \powerset{\powerset{(\{x\} \cup B)}}.$$
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Since this holds for all sets $t$,
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$\{x\} \times B \subseteq \powerset{\powerset{(\{x\} \cup B)}}$.
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\paragraph{($\Rightarrow$)}%
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\paragraph{(ii)}%
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\label{par:exercise-5.5a-ii}
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Suppose $y \in C$.
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Then there exists some $a \in A$ such that
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$$\forall x, \left[ x \in y \iff
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\exists b \in B, x = \left< a, b \right> \right].$$
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By the \nameref{ref:extensionality-axiom},
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\begin{align*}
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y
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& = \{ \left< a, b \right> \mid b \in B \} \\
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& = \{ \left< x, b \right> \mid x \in \{a\} \land b \in B \} \\
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& = \{ \{a\} \times B \}.
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\end{align*}
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By the \nameref{ref:subset-axioms}, there exists a set $C$ such that for any
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set $y$, $$y \in C \iff
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y \in \powerset{\powerset{(\{x\} \cup B)}} \land
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(\exists b \in B, y = \left< x, b \right>).$$
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The above equation and \eqref{sub:exercise-5.5a-eq1} implies $C$ contains
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only ordered pairs of the desired form.
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By \nameref{par:exercise-5.5a-i}, $C$ contains them all.
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\paragraph{($\Leftarrow$)}%
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\paragraph{Conclusion}%
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Since $x$ was an arbitrarily chosen member of $A$,
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\nameref{par:exercise-5.5a-ii} immediately implies that
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$\{\{x\} \times B \mid x \in A\}$ is indeed a set.
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Suppose $y = \{a\} \times B$ for some $a \in A$.
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By \nameref{sub:cartesian-product}, $x \in \{a\} \times B$ if and only if
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$\exists b \in B$ such that $x = \left< a, b \right>$.
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But then $x \in y$ if and only if $\exists b \in B$ such that
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$x = \left< a, b \right>$.
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This immediately proves $y \in C$.
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\end{proof}
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@ -2633,4 +2625,87 @@ With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$.
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\end{proof}
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\section{Relations}%
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\label{sec:relations}
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\subsection{\verified{Theorem 3D}}%
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\label{sub:theorem-3d}
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\begin{theorem}[3D]
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If $\left< x, y \right> \in A$, then $x$ and $y$ belong to $\bigcup\bigcup A$.
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\end{theorem}
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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\_3d}
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Let $A$ be a set and $\left< x, y \right> \in A$.
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By definition of an \nameref{ref:ordered-pair},
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$$\left< x, y \right> = \{\{x\}, \{x, y\}\}.$$
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By \nameref{sub:exercise-3.3}, $\{\{x\}, \{x, y\}\} \subseteq \bigcup A$.
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Then $\{x, y\} \in \bigcup A$.
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Another application of \nameref{sub:exercise-3.3} implies
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$\{x, y\} \subseteq \bigcup\bigcup A$.
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Therefore $x, y \in \bigcup\bigcup A$.
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\end{proof}
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\section{Exercises 6}%
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\label{sec:exercises-6}
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\subsection{\unverified{Exercise 6.6}}%
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\label{sub:exercise-6.6}
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Show that a set $A$ is a relation iff
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$A \subseteq \textop{dom} A \times \textop{ran} A$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 6.7}}%
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\label{sub:exercise-6.7}
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Show that if $R$ is a relation, then $\textop{fld} R = \bigcup\bigcup R$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 6.8}}%
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\label{sub:exercise-6.8}
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Show that for any set $\mathscr{A}$:
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\begin{align*}
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\textop{dom} \bigcup{\mathscr{A}}
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& = \bigcup\; \{ \textop{dom} R \mid R \in \mathscr{A} \},
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\textop{ran} \bigcup{\mathscr{A}}
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& = \bigcup\; \{ \textop{ran} R \mid R \in \mathscr{A} \}.
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\end{align*}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Exercise 6.9}}%
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\label{sub:exercise-6.9}
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Discuss the result of replacing the union operation by the intersection
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operation in the preceding problem.
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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@ -1,5 +1,6 @@
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import Mathlib.Data.Set.Basic
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import Bookshelf.Enderton.Set.Chapter_2
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import Common.Logic.Basic
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import Common.Set.Basic
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@ -221,7 +222,61 @@ In other words, show that `{{x} × B | x ∈ A}` is a set.
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theorem exercise_5_5a {A : Set α} {B : Set β}
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: ∃ C : Set (Set (α × β)),
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y ∈ C ↔ ∃ x ∈ A, y = Set.prod {x} B := by
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sorry
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let C := {y ∈ 𝒫 (Set.prod A B) | ∃ a ∈ A, ∀ x, (x ∈ y ↔ ∃ b ∈ B, x = (a, b))}
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refine ⟨C, ?_⟩
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apply Iff.intro
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· intro hC
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simp only [Set.mem_setOf_eq] at hC
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have ⟨_, ⟨a, ⟨ha, h⟩⟩⟩ := hC
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refine ⟨a, ⟨ha, ?_⟩⟩
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ext x
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apply Iff.intro
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· intro hxy
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unfold Set.prod
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simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
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have ⟨b, ⟨hb, hx⟩⟩ := (h x).mp hxy
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rw [Prod.ext_iff] at hx
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simp only at hx
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rw [← hx.right] at hb
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exact ⟨hx.left, hb⟩
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· intro hx
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simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hx
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have := (h (a, x.snd)).mpr ⟨x.snd, ⟨hx.right, rfl⟩⟩
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have hxab : x = (a, x.snd) := by
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ext <;> simp
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exact hx.left
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rwa [← hxab] at this
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· intro ⟨x, ⟨hx, hy⟩⟩
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show y ∈ 𝒫 Set.prod A B ∧ ∃ a, a ∈ A ∧
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∀ (x : α × β), x ∈ y ↔ ∃ b, b ∈ B ∧ x = (a, b)
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apply And.intro
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· simp only [Set.mem_powerset_iff]
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rw [hy]
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unfold Set.prod
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simp only [
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Set.mem_singleton_iff,
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Set.setOf_subset_setOf,
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and_imp,
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Prod.forall
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]
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intro a b ha hb
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exact ⟨by rw [ha]; exact hx, hb⟩
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· refine ⟨x, ⟨hx, ?_⟩⟩
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intro p
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apply Iff.intro
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· intro hab
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rw [hy] at hab
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unfold Set.prod at hab
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simp only [Set.mem_singleton_iff, Set.mem_setOf_eq] at hab
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exact ⟨p.2, ⟨hab.right, by ext; exact hab.left; simp⟩⟩
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· intro ⟨b, ⟨hb, hab⟩⟩
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rw [hy]
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unfold Set.prod
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simp only [Set.mem_singleton_iff, Set.mem_setOf_eq]
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rw [Prod.ext_iff] at hab
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simp only at hab
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rw [hab.right]
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exact ⟨hab.left, hb⟩
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/-- ### Exercise 5.5b
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@ -258,4 +313,19 @@ theorem exercise_5_5b {A : Set α} (B : Set β)
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rw [← ha] at h
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exact ⟨h, hb⟩
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/-- ### Theorem 3D
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If `⟨x, y⟩ ∈ A`, then `x` and `y` belong to `⋃ ⋃ A`.
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-/
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theorem theorem_3d {A : Set (Set (Set (α ⊕ α)))} (h : OrderedPair x y ∈ A)
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: Sum.inl x ∈ ⋃₀ (⋃₀ A) ∧ Sum.inr y ∈ ⋃₀ (⋃₀ A) := by
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have hp : OrderedPair x y ⊆ ⋃₀ A := Chapter_2.exercise_3_3 (OrderedPair x y) h
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have hp' : ∀ t, t ∈ {{Sum.inl x}, {Sum.inl x, Sum.inr y}} → t ∈ ⋃₀ A := hp
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have hq := hp' {Sum.inl x, Sum.inr y} (by simp)
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have hq' := Chapter_2.exercise_3_3 {Sum.inl x, Sum.inr y} hq
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have : ∀ t, t ∈ {Sum.inl x, Sum.inr y} → t ∈ ⋃₀ (⋃₀ A) := hq'
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exact ⟨this (Sum.inl x) (by simp), this (Sum.inr y) (by simp)⟩
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end Enderton.Set.Chapter_3
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@ -134,6 +134,7 @@
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\newcommand{\ioc}[2]{\left(#1, #2\right]}
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\newcommand{\ioo}[2]{\left(#1, #2\right)}
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\newcommand{\powerset}[1]{\mathscr{P}#1}
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\newcommand{\textop}[1]{\mathop{\text{#1}}}
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\newcommand{\ubar}[1]{\text{\b{$#1$}}}
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\let\oldemptyset\emptyset
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