Enderton. Finish function theorems.
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@ -9,6 +9,7 @@
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\newcommand{\dom}[1]{\textop{dom}{#1}}
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\newcommand{\fld}[1]{\textop{fld}{#1}}
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\newcommand{\ran}[1]{\textop{ran}{#1}}
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\newcommand{\img}[2]{#1\left\llbracket#2\right\rrbracket}
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\begin{document}
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@ -125,7 +126,7 @@ One-to-one functions are sometimes called \textbf{injections}.
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Let $A$ and $F$ be arbitrary sets.
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The \textbf{image of $A$ under $F$} is the set
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\begin{align*}
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F[A]
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\img{F}{A}
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& = \ran{(F \restriction A)} \\
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& = \{v \mid (\exists u \in A) uFv\}.
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\end{align*}
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@ -3413,59 +3414,233 @@ 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{\unverified{Theorem 3K}}%
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\label{sub:theorem-3k}
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\subsection{\partial{Theorem 3K(a)}}%
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\label{sub:theorem-3k-a}
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\begin{theorem}[3K]
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\begin{theorem}[3K(a)]
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The following hold for any sets. ($F$ need not be a function.)
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\begin{enumerate}[(a)]
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\item The image of a union is the union of the images:
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$$F[A \cup B] = F[A] \cup F[B] \quad\text{and}\quad
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F\left[\bigcup{\mathscr{A}}\right] =
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\bigcup\;\{F[A] \mid A \in \mathscr{A}\}.$$
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\item The image of an intersection is included in the intersection of the
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images:
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$$F[A \cap B] \subseteq F[A] \cap F[B] \quad\text{and}\quad
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F\left[\bigcap\mathscr{A}\right] \subseteq
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\bigcap\;\{F[A] \mid A \in \mathscr{A}\}$$
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for nonempty $\mathscr{A}$.
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Equality holds if $F$ is single-rooted.
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\item The image of a difference includes the difference of the images:
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$$F[A] - F[B] \subseteq F[A - B].$$
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Equality holds if $F$ is single-rooted.
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\end{enumerate}
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The image of a union is the union of the images:
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\begin{equation}
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\label{sub:theorem-3k-a-eq1}
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\img{F}{A \cup B} = \img{F}{A} \cup \img{F}{B}
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\end{equation}
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and
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\begin{equation}
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\label{sub:theorem-3k-a-eq2}
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\img{F}{\bigcup{\mathscr{A}}} =
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\bigcup\;\{\img{F}{A} \mid A \in \mathscr{A}\}.
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\end{equation}
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\end{theorem}
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\begin{proof}
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TODO
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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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\paragraph{(i)}%
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By definition of the \nameref{ref:image} of a set:
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\begin{align*}
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\img{F}{A \cup B}
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& = \{v \mid \exists u, u \in A \cup B \land uFv\} \\
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& = \{v \mid \exists u,
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(u \in A \land uFv) \lor (u \in B \land uFv)\} \\
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& = \{v \mid (\exists u \in A) uFv\} \cup
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\{v \mid (\exists u \in B) uFv\} \\
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& = \img{F}{A} \cup \img{F}{B}.
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\end{align*}
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\paragraph{(ii)}%
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We prove that both sides of \eqref{sub:theorem-3k-a-eq2} is a subset of the
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other.
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\subparagraph{($\subseteq$)}%
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Let $v \in \img{F}{\bigcup{\mathscr{A}}}$.
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By definition of the \nameref{ref:image} of a set, there exists a set $u$
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such that $u \in \bigcup{\mathscr{A}} \land uFv$.
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Then, by definition of the union of sets, there exists some
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$A \in \mathscr{A}$ such that $u \in A$.
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Therefore $v \in \img{F}{A}$ meaning
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$v \in \bigcup\{\img{F}{A} \mid A \in \mathscr{A}\}$.
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\subparagraph{($\supseteq$)}%
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Let $v \in \bigcup\{\img{F}{A} \mid A \in \mathscr{A}\}$.
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Then there exists some $b \in \{\img{F}{A} \mid A \in \mathscr{A}\}$ such
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that $v \in b$.
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In other words, there exists some $A \in \mathscr{A}$ such that
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$v \in b = \img{F}{A}$.
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By definition of the \nameref{ref:image} of a set, there exists a set $u$
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such that $u \in A \land uFv$.
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But this implies that $u \in \bigcup{\mathscr{A}} \land uFv$.
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Therefore $v \in \img{F}{\bigcup{\mathscr{A}}}$.
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\end{proof}
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\subsection{\unverified{Corollary 3L}}%
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\subsection{\partial{Theorem 3K(b)}}%
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\label{sub:theorem-3k-b}
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\begin{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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\begin{equation}
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\label{sub:theorem-3k-b-eq1}
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\img{F}{A \cap B} \subseteq \img{F}{A} \cap \img{F}{B}
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\end{equation}
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and
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\begin{equation}
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\label{sub:theorem-3k-b-eq2}
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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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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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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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\eqref{sub:theorem-3k-b-eq1} and (iv) \eqref{sub:theorem-3k-b-eq2} hold
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under equality.
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\paragraph{(i)}%
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\label{par:theorem-3k-b-i}
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Let $v \in \img{F}{A \cap B}$.
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By definition of the \nameref{ref:image} of a set,
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$\exists u \in A \cap B, uFv$.
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Then $u \in A \land uFv$ and $u \in B \land uFv$.
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Therefore $v \in \img{F}{A} \cap \img{F}{B}$.
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\paragraph{(ii)}%
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\label{par:theorem-3k-b-ii}
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Let $v \in \img{F}{\bigcap{\mathscr{A}}}$.
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By definition of the \nameref{ref:image} of a set,
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$\exists u \in \bigcap{\mathscr{A}}, uFv$.
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Then $\exists u, (\forall A \in \mathscr{A}, u \in A) \land uFv$.
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This implies that $\forall A \in \mathscr{A}, \exists u \in A, uFv$.
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Then $\forall A \in \mathscr{A}, v \in \img{F}{A}$.
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Thus $v \in \bigcap\{\img{F}{A} \mid A \in \mathscr{A}\}$.
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\paragraph{(iii)}%
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Suppose $F$ is single-rooted.
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By \nameref{par:theorem-3k-b-i},
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$$\img{F}{A \cap B} \subseteq \img{F}{A} \cap \img{F}{B}.$$
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All that remains is showing
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$$\img{F}{A} \cap \img{F}{B} \subseteq \img{F}{A \cap B}.$$
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Let $v \in \img{F}{A} \cap \img{F}{B}$.
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Then $v \in \img{F}{A}$ and $v \in \img{F}{B}$.
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That is, $\exists u \in A, uFv$ and $\exists w \in B, wFv$.
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Since $F$ is single rooted, it follows $u = w$.
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Thus $u \in A \cap B \land uFv$ meaning $v \in \img{F}{A \cap B}$.
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\paragraph{(iv)}%
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Suppose $F$ is single-rooted.
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By \nameref{par:theorem-3k-b-ii},
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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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All that remains is showing
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$$\bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\} \subseteq
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\img{F}{\bigcap\mathscr{A}}.$$
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Let $v \in \bigcap\;\{\img{F}{A} \mid A \in \mathscr{A}\}$.
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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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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{\partial{Theorem 3K(c)}}%
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\label{sub:theorem-3k-c}
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\begin{theorem}[3K(c)]
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The following hold for any sets. ($F$ need not be a function.)
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The image of a difference includes the difference of the images:
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\begin{equation}
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\label{sub:theorem-3k-c-eq1}
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\img{F}{A} - \img{F}{B} \subseteq \img{F}{A - B}.
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\end{equation}
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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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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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\paragraph{(i)}%
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\label{par:theorem-3k-c-i}
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Let $v \in \img{F}{A} - \img{F}{B}$.
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By definition of the difference of two sets,
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$v \in \img{F}{A}$ and $v \not\in \img{F}{B}$.
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By definition of the \nameref{ref:image} of a set, there exists a set
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$u \in A$ such that $\left< u, v \right> \in F$.
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Likewise, $\forall w \in B, \left< w, v \right> \not\in F$.
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Thus $u \not\in B$, since otherwise we get an immediate contradiction.
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Therefore $u \in A - B$ meaning $v \in \img{F}{A - B}$.
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\paragraph{(ii)}%
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Suppose $F$ is single-rooted.
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By \nameref{par:theorem-3k-c-i},
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$$\img{F}{A} - \img{F}{B} \subseteq \img{F}{A - B}.$$
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All that remains is showing
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$$\img{F}{A - B} \subseteq \img{F}{A} - \img{F}{B}.$$
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Let $v \in \img{F}{A - B}$.
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By definition of the \nameref{ref:image} of a set, there exists a set
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$u \in A - B$ such that $uFv$.
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Then $u \in A$ and $u \not\in B$.
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The former membership relation implies $v \in \img{F}{A}$.
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The latter implies $v \not\in \img{F}{B}$ since $F$ being single-rooted
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would otherwise invoke an immediate contradiction.
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Thus $v \in \img{F}{A} - \img{F}{B}$.
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\end{proof}
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\subsection{\partial{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[\bigcup\mathscr{A}\right]
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& = \bigcup\;\{G^{-1}[A] \mid A \in \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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\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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\text{ for } \mathscr{A} \neq \emptyset, \\
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G^{-1}[A - B]
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& = G^{-1}[A] - G^{-1}[B].
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\end{align*}
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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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\label{sub:corollary-3l-eq3}
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\end{align}
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\end{theorem}
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\begin{proof}
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TODO
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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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Likewise \nameref{sub:theorem-3k-c} implies \eqref{sub:corollary-3l-eq3}.
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\end{proof}
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@ -8,6 +8,7 @@
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\usepackage{fontawesome5}
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\usepackage{mathrsfs}
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\usepackage{soul}
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\usepackage{stmaryrd}
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\usepackage[usenames,dvipsnames]{xcolor}
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% `hyperref` comes after `xr-hyper`.
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\usepackage{xr-hyper}
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