Enderton. Exercise 3.31, Theorem 3K, and corollary.
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@ -44,8 +44,18 @@ For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
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\end{axiom}
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\section{\pending{Cartesian Product}}%
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\label{ref:cartesian-product}
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Let $I$ be a set and let $H$ be a \nameref{ref:function} whose domain includes $I$.
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Then for each $i$ in $I$ we have the set $H(i)$.
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We define the \textbf{cartesian product} of the $H(i)$'s as
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$$\bigtimes_{i \in I} H(i) = \{f \mid
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f \text{ is a function with domain } I \text{ and }
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(\forall i \in I) f(i) \in H(i)\}.$$
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\section{\defined{Compatible}}%
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\label{sec:compatible}
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\label{ref:compatible}
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A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and
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only if for all $x$ and $y$ in $A$,
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@ -2569,9 +2579,8 @@ If not, then under what conditions does equality hold?
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\end{proof}
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\subsection{\verified{Cartesian Product}}%
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\subsection{\verified{Corollary 3C}}%
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\label{sub:corollary-3c}
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\label{sub:cartesian-product}
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\begin{theorem}[3C]
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@ -3018,7 +3027,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{proof}
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\subsection{\pending{Theorem 3K(a)}}%
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\subsection{\verified{Theorem 3K(a)}}%
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\label{sub:theorem-3k-a}
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\begin{theorem}[3K(a)]
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@ -3040,6 +3049,9 @@ 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\_3k\_a}
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Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets.
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We prove (i) \eqref{sub:theorem-3k-a-eq1} and (ii)
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\eqref{sub:theorem-3k-a-eq2}.
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@ -3086,7 +3098,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{proof}
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\subsection{\pending{Theorem 3K(b)}}%
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\subsection{\verified{Theorem 3K(b)}}%
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\label{sub:theorem-3k-b}
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\begin{theorem}[3K(b)]
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@ -3103,13 +3115,23 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\img{F}{\bigcap\mathscr{A}} \subseteq
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\bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}.
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\end{equation}
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for nonempty $\mathscr{A}$.
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Equality holds if $F$ is single-rooted.
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\end{theorem}
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\begin{proof}
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Let $F$, $A$, $B$, and $\mathscr{A}$ be arbitrary sets.
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\statementpadding
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.theorem\_3k\_b\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.theorem\_3k\_b\_ii}
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Let $F$, $A$, $B$ be arbitrary sets.
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Let $\mathscr{A}$ be a nonempty set.
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We first prove (i) \eqref{sub:theorem-3k-b-eq1} and (ii)
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\eqref{sub:theorem-3k-b-eq2}.
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Then, assuming $F$ is single-rooted, we prove both (iii)
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@ -3162,14 +3184,14 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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Then $\forall A \in \mathscr{A}, v \in \img{F}{A}$.
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By definition of the \nameref{ref:image} of a set,
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$\forall A \in \mathscr{A}, \exists u \in A, uFv$.
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Since $F$ is single-rooted, it follows that
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$\exists u, \forall A \in \mathscr{A}, u \in A \land uFv$.
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Since $F$ is single-rooted and $\mathscr{A}$ is nonempty, it follows that
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$\exists u, (\forall A \in \mathscr{A}, u \in A) \land uFv$.
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Equivalently, $\exists u \in \bigcap{A}, uFv$.
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Thus $v \in \img{F}{\bigcap{A}}$.
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\end{proof}
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\subsection{\pending{Theorem 3K(c)}}%
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\subsection{\verified{Theorem 3K(c)}}%
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\label{sub:theorem-3k-c}
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\begin{theorem}[3K(c)]
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@ -3186,6 +3208,14 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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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.theorem\_3k\_c\_i}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.theorem\_3k\_c\_ii}
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We prove that (i) \eqref{sub:theorem-3k-c-eq1} holds and (ii) equality holds
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if $F$ is single-rooted.
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@ -3219,21 +3249,21 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{proof}
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\subsection{\pending{Corollary 3L}}%
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\subsection{\verified{Corollary 3L}}%
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\label{sub:corollary-3l}
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\begin{theorem}[3L]
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For any function $G$ and sets $A$, $B$, and $\mathscr{A}$:
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\begin{align}
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G^{-1}\left\llbracket\bigcup\mathscr{A}\right\rrbracket
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& = \bigcup\;\{G^{-1}[A] \mid A \in \mathscr{A}\},
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\img{G^{-1}}{\bigcup{\mathscr{A}}}
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& = \bigcup\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\},
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\label{sub:corollary-3l-eq1} \\
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G^{-1}\left[\bigcap\mathscr{A}\right]
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& = \bigcap\;\{G^{-1}[A] \mid A \in \mathscr{A}\}
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\img{G^{-1}}{\bigcap{\mathscr{A}}}
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& = \bigcap\;\{\img{G^{-1}}{A} \mid A \in \mathscr{A}\}
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\text{ for } \mathscr{A} \neq \emptyset,
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\label{sub:corollary-3l-eq2} \\
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G^{-1}[A - B] & = G^{-1}[A] - G^{-1}[B].
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\img{G^{-1}}{A - B} & = \img{G^{-1}}{A} - \img{G^{-1}}{B}.
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\label{sub:corollary-3l-eq3}
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\end{align}
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@ -3241,6 +3271,17 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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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.corollary\_3l\_i}
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\lean*{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.corollary\_3l\_ii}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.corollary\_3l\_iii}
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\nameref{sub:theorem-3k-a} implies \eqref{sub:corollary-3l-eq1}.
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Because the inverse of a function is always single-rooted,
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\nameref{sub:theorem-3k-b} implies \eqref{sub:corollary-3l-eq2}.
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@ -3441,7 +3482,7 @@ Show that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
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{Enderton.Set.Chapter\_3.exercise\_3\_2a}
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Let $A$, $B$, and $C$ be arbitrary sets.
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Then by definition of the \nameref{sub:cartesian-product} and union of sets,
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Then by \nameref{sub:corollary-3c} and the definition of the union of sets,
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\begin{align*}
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A \times (B \cup C)
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& = \{ \left< x, y \right> \mid x \in A \land y \in (B \cup C) \} \\
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@ -3467,7 +3508,7 @@ Show that if $A \times B = A \times C$ and $A \neq \emptyset$, then $B = C$.
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{Enderton.Set.Chapter\_3.exercise\_3\_2b}
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Let $A$, $B$, and $C$ be arbitrary sets such that $A \neq \emptyset$.
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By definition of the \nameref{sub:cartesian-product},
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By \nameref{sub:corollary-3c},
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\begin{align}
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A \times B & = \{ \left< x, y \right> \mid x \in A \land y \in B \}
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& \label{sub:exercise-3.2b-eq1} \\
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@ -3513,7 +3554,7 @@ Show that $A \times \bigcup \mathscr{B} =
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{Enderton.Set.Chapter\_3.exercise\_3\_3}
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Let $A$ and $\mathscr{B}$ be arbitrary sets.
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By definition of the \nameref{sub:cartesian-product} and the union of sets,
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By \nameref{sub:corollary-3c} and the definition of the union of sets,
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\begin{align*}
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A \times \bigcup\mathscr{B}
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& = \{ \left< x, y \right> \mid
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@ -3564,10 +3605,10 @@ In other words, show that $\{\{x\} \times B \mid x \in A\}$ 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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By \nameref{sub:corollary-3c}, $\{a\} \times B$ is a set.
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Again by the \nameref{ref:pairing-axiom}, $\{\{a\} \times B\}$ is a set.
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Next, by another application of \nameref{sub:cartesian-product}, $A \times B$
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Next, by another application of \nameref{sub:corollary-3c}, $A \times B$
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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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@ -3593,7 +3634,7 @@ In other words, show that $\{\{x\} \times B \mid x \in A\}$ is a set.
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\paragraph{($\Leftarrow$)}%
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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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By \nameref{sub:corollary-3c}, $x \in \{a\} \times B$ 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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@ -3618,7 +3659,7 @@ With $A$, $B$, and $C$ as above, show that $A \times B = \bigcup C$.
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A \times B = \bigcup\; \{\{x\} \times B \mid x \in A\}.
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\end{equation}
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The left-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of
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\nameref{sub:cartesian-product}.
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\nameref{sub:corollary-3c}.
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The right-hand side of \eqref{sub:exercise-3.5b-eq1} is a set by virtue of
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\nameref{sub:exercise-3.5a}.
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We prove the set on each side is a subset of the other.
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@ -4193,9 +4234,8 @@ Show that $F \restriction A = F \cap (A \times \ran{F})$.
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\begin{proof}
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Let $F$ and $A$ be arbitrary sets.
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By definition of the \nameref{ref:restriction}, intersection,
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\nameref{ref:range}, and \nameref{sub:cartesian-product} of sets,
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Then
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By \nameref{sub:corollary-3c} and definition of the \nameref{ref:restriction},
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intersection, and \nameref{ref:range} of sets,
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\begin{align*}
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F \restriction A
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& = \{\left< u, v \right> \mid uFv \land u \in A\} \\
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@ -4559,7 +4599,7 @@ Define
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\end{proof}
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\subsection{\sorry{Exercise 3.31}}%
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\subsection{\unverified{Exercise 3.31}}%
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\label{sub:exercise-3.31}
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Show that from the first form of the axiom of choice we can prove the second
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@ -4567,7 +4607,55 @@ Show that from the first form of the axiom of choice we can prove the second
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\begin{proof}
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TODO
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We prove the first form holds if and only if the second form holds.
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\paragraph{($\Rightarrow$)}%
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We assume the first form of the axiom of choice.
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Let $I$ be a set and $H$ be a function with $\dom{H} = I$.
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Furthermore, suppose $H(i) \neq \emptyset$ for all $i \in I$.
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By definition of the \nameref{ref:cartesian-product},
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$$\bigtimes_{i \in I} H(i) = \{f \mid
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f \text{ is a function with } \dom{f} = I \text{ and }
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(\forall i \in I) f(i) \in H(i)\}.$$
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Consider the relation $R$ formed by
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$$R = \bigcup_{i \in I} \{i\} \times H(i).$$
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By the \nameref{ref:axiom-of-choice-1}, there exists a function
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$f \subseteq R$ with $\dom{f} = I$.
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Furthermore, for all $i \in I$, it must be $f(i) \in H(i)$ by construction.
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Then $f$ is a member of $\bigtimes_{i \in I} H(i)$.
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That is, $\bigtimes_{i \in I} H(i) \neq \emptyset$.
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\paragraph{($\Leftarrow$)}%
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We assume the second form of the axiom of choice.
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Let $R$ be an arbitrary relation.
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There are two cases to consider:
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\subparagraph{Case 1}%
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Suppose $\ran{R} = \emptyset$.
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Then $R = \emptyset$.
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Thus the function $\emptyset \subseteq R$ satisfies
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$\dom{\emptyset} = \dom{R}$.
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\subparagraph{Case 2}%
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Suppose $\ran{R} \neq \emptyset$.
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Let $I = \dom{R}$ and define $H \colon I \rightarrow \{\ran{R}\}$ as
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$H(i) = \ran{R}$ for all $i \in I$.
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By the \nameref{ref:axiom-of-choice-2},
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$\bigtimes_{i \in I} H(i) \neq \emptyset$.
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By definition of the \nameref{ref:cartesian-product}, there exists some
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function $f$ such that $\dom{f} = I$ and
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$(\forall i \in I) f(i) \in H(i) = \ran{R}$.
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Thus $\dom{f} = \dom{R}$ and $f \subseteq R$ as desired.
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\paragraph{Conclusion}%
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The above cases are exhaustive and yield the same conclusion: for any
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relation $R$ there exists a function $f \subseteq R$ such that
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$\dom{f} = \dom{R}$.
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\end{proof}
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@ -534,6 +534,186 @@ theorem theorem_3j_b {F : Set.Relation α} {A B : Set α}
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(∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by
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sorry
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/-- #### Theorem 3K (a)
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The following hold for any sets. (`F` need not be a function.)
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The image of a union is the union of the images:
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```
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F⟦⋃ 𝓐⟧ = ⋃ {F⟦A⟧ | A ∈ 𝓐}
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```
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-/
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theorem theorem_3k_a {F : Set.Relation α} {𝓐 : Set (Set α)}
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: F.image (⋃₀ 𝓐) = ⋃₀ { F.image A | A ∈ 𝓐 } := by
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rw [Set.Subset.antisymm_iff]
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apply And.intro
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· show ∀ v, v ∈ F.image (⋃₀ 𝓐) → v ∈ ⋃₀ { F.image A | A ∈ 𝓐 }
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intro v hv
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unfold image at hv
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simp only [Set.mem_sUnion, Set.mem_setOf_eq] at hv
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have ⟨u, hu⟩ := hv
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have ⟨A, hA⟩ := hu.left
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simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and]
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refine ⟨A, hA.left, ?_⟩
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show v ∈ F.image A
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unfold image
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simp only [Set.mem_setOf_eq]
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exact ⟨u, hA.right, hu.right⟩
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· show ∀ v, v ∈ ⋃₀ {x | ∃ A, A ∈ 𝓐 ∧ F.image A = x} → v ∈ F.image (⋃₀ 𝓐)
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intro v hv
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simp only [Set.mem_sUnion, Set.mem_setOf_eq, exists_exists_and_eq_and] at hv
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have ⟨A, hA⟩ := hv
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unfold image at hA
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simp only [Set.mem_setOf_eq] at hA
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have ⟨u, hu⟩ := hA.right
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unfold image
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simp only [Set.mem_sUnion, Set.mem_setOf_eq]
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exact ⟨u, ⟨A, hA.left, hu.left⟩, hu.right⟩
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/-! #### Theorem 3K (b)
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The following hold for any sets. (`F` need not be a function.)
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The image of an intersection is included in the intersection of the images:
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```
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F⟦⋂ 𝓐⟧ ⊆ ⋂ {F⟦A⟧ | A ∈ 𝓐}
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```
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Equality holds if `F` is single-rooted.
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-/
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theorem theorem_3k_b_i {F : Set.Relation α} {𝓐 : Set (Set α)}
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: F.image (⋂₀ 𝓐) ⊆ ⋂₀ { F.image A | A ∈ 𝓐} := by
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show ∀ v, v ∈ F.image (⋂₀ 𝓐) → v ∈ ⋂₀ { F.image A | A ∈ 𝓐}
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intro v hv
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unfold image at hv
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simp only [Set.mem_sInter, Set.mem_setOf_eq] at hv
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have ⟨u, hu⟩ := hv
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simp only [
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Set.mem_sInter,
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Set.mem_setOf_eq,
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forall_exists_index,
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and_imp,
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forall_apply_eq_imp_iff₂
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]
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intro A hA
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unfold image
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simp only [Set.mem_setOf_eq]
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exact ⟨u, hu.left A hA, hu.right⟩
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theorem theorem_3k_b_ii {F : Set.Relation α} {𝓐 : Set (Set α)}
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(hF : F.isSingleRooted) (h𝓐 : Set.Nonempty 𝓐)
|
||||
: F.image (⋂₀ 𝓐) = ⋂₀ { F.image A | A ∈ 𝓐} := by
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
refine ⟨theorem_3k_b_i, ?_⟩
|
||||
show ∀ v, v ∈ ⋂₀ {x | ∃ A, A ∈ 𝓐 ∧ image F A = x} → v ∈ image F (⋂₀ 𝓐)
|
||||
intro v hv
|
||||
simp only [
|
||||
Set.mem_sInter,
|
||||
Set.mem_setOf_eq,
|
||||
forall_exists_index,
|
||||
and_imp,
|
||||
forall_apply_eq_imp_iff₂
|
||||
] at hv
|
||||
unfold image at hv
|
||||
simp only [Set.mem_setOf_eq] at hv
|
||||
have ⟨u, hu⟩ : ∃ u, (∀ (a : Set α), a ∈ 𝓐 → u ∈ a) ∧ (u, v) ∈ F := by
|
||||
have ⟨A, hA⟩ := h𝓐
|
||||
have ⟨_, ⟨_, hv'⟩⟩ := hv A hA
|
||||
have ⟨u, hu⟩ := hF v (mem_pair_imp_snd_mem_ran hv')
|
||||
simp only [and_imp] at hu
|
||||
refine ⟨u, ?_, hu.left.right⟩
|
||||
intro a ha
|
||||
have ⟨u₁, hu₁⟩ := hv a ha
|
||||
have := hu.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
|
||||
rw [← this]
|
||||
exact hu₁.left
|
||||
unfold image
|
||||
simp only [Set.mem_sInter, Set.mem_setOf_eq]
|
||||
exact ⟨u, hu⟩
|
||||
|
||||
/-! #### Theorem 3K (c)
|
||||
|
||||
The following hold for any sets. (`F` need not be a function.)
|
||||
The image of a difference includes the difference of the images:
|
||||
```
|
||||
F⟦A⟧ - F⟦B⟧ ⊆ F⟦A - B⟧.
|
||||
```
|
||||
Equality holds if `F` is single-rooted.
|
||||
-/
|
||||
|
||||
theorem theorem_3k_c_i {F : Set.Relation α} {A B : Set α}
|
||||
: F.image A \ F.image B ⊆ F.image (A \ B) := by
|
||||
show ∀ v, v ∈ F.image A \ F.image B → v ∈ F.image (A \ B)
|
||||
intro v hv
|
||||
have hv' : v ∈ image F A ∧ v ∉ image F B := hv
|
||||
conv at hv' => arg 1; unfold image; simp only [Set.mem_setOf_eq, eq_iff_iff]
|
||||
have ⟨u, hu⟩ := hv'.left
|
||||
have hw : ∀ w ∈ B, (w, v) ∉ F := by
|
||||
intro w hw nw
|
||||
have nv := hv'.right
|
||||
unfold image at nv
|
||||
simp only [Set.mem_setOf_eq, not_exists, not_and] at nv
|
||||
exact absurd nw (nv w hw)
|
||||
have hu' : u ∉ B := by
|
||||
by_contra nu
|
||||
exact absurd hu.right (hw u nu)
|
||||
unfold image
|
||||
simp only [Set.mem_diff, Set.mem_setOf_eq]
|
||||
exact ⟨u, ⟨hu.left, hu'⟩, hu.right⟩
|
||||
|
||||
theorem theorem_3k_c_ii {F : Set.Relation α} {A B : Set α}
|
||||
(hF : F.isSingleRooted)
|
||||
: F.image A \ F.image B = F.image (A \ B) := by
|
||||
rw [Set.Subset.antisymm_iff]
|
||||
refine ⟨theorem_3k_c_i, ?_⟩
|
||||
show ∀ v, v ∈ image F (A \ B) → v ∈ image F A \ image F B
|
||||
intro v hv
|
||||
unfold image at hv
|
||||
simp only [Set.mem_diff, Set.mem_setOf_eq] at hv
|
||||
have ⟨u, hu⟩ := hv
|
||||
have hv₁ : v ∈ F.image A := by
|
||||
unfold image
|
||||
simp only [Set.mem_setOf_eq]
|
||||
exact ⟨u, hu.left.left, hu.right⟩
|
||||
have hv₂ : v ∉ F.image B := by
|
||||
intro nv
|
||||
unfold image at nv
|
||||
simp only [Set.mem_setOf_eq] at nv
|
||||
have ⟨u₁, hu₁⟩ := nv
|
||||
have ⟨x, hx⟩ := hF v (mem_pair_imp_snd_mem_ran hu.right)
|
||||
simp only [and_imp] at hx
|
||||
have hr₁ := hx.right u (mem_pair_imp_fst_mem_dom hu.right) hu.right
|
||||
have hr₂ := hx.right u₁ (mem_pair_imp_fst_mem_dom hu₁.right) hu₁.right
|
||||
rw [hr₂, ← hr₁] at hu₁
|
||||
exact absurd hu₁.left hu.left.right
|
||||
exact ⟨hv₁, hv₂⟩
|
||||
|
||||
/-! #### Corollary 3L
|
||||
|
||||
For any function `G` and sets `A`, `B`, and `𝓐`:
|
||||
|
||||
```
|
||||
G⁻¹⟦⋃ 𝓐⟧ = ⋃ {G⁻¹⟦A⟧ | A ∈ 𝓐},
|
||||
G⁻¹⟦𝓐⟧ = ⋂ {G⁻¹⟦A⟧ | A ∈ 𝓐} for 𝓐 ≠ ∅,
|
||||
G⁻¹⟦A - B⟧ = G⁻¹⟦A⟧ - G⁻¹⟦B⟧.
|
||||
```
|
||||
-/
|
||||
|
||||
theorem corollary_3l_i {G : Set.Relation α} {𝓐 : Set (Set α)}
|
||||
: G.inv.image (⋃₀ 𝓐) = ⋃₀ {G.inv.image A | A ∈ 𝓐} := theorem_3k_a
|
||||
|
||||
theorem corollary_3l_ii {G : Set.Relation α} {𝓐 : Set (Set α)}
|
||||
(hG : G.isSingleValued) (h𝓐 : Set.Nonempty 𝓐)
|
||||
: G.inv.image (⋂₀ 𝓐) = ⋂₀ {G.inv.image A | A ∈ 𝓐} := by
|
||||
have hG' : G.inv.isSingleRooted :=
|
||||
single_valued_self_iff_single_rooted_inv.mp hG
|
||||
exact theorem_3k_b_ii hG' h𝓐
|
||||
|
||||
theorem corollary_3l_iii {G : Set.Relation α} {A B : Set α}
|
||||
(hG : G.isSingleValued)
|
||||
: G.inv.image (A \ B) = G.inv.image A \ G.inv.image B := by
|
||||
have hG' : G.inv.isSingleRooted :=
|
||||
single_valued_self_iff_single_rooted_inv.mp hG
|
||||
exact (theorem_3k_c_ii hG').symm
|
||||
|
||||
end
|
||||
|
||||
end Enderton.Set.Chapter_3
|
||||
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Reference in New Issue