Enderton. Finish function theorems.

finite-set-exercises
Joshua Potter 2023-06-28 13:19:59 -06:00
parent 62f259358d
commit 76d294a47f
2 changed files with 207 additions and 31 deletions

View File

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

View File

@ -8,6 +8,7 @@
\usepackage{fontawesome5}
\usepackage{mathrsfs}
\usepackage{soul}
\usepackage{stmaryrd}
\usepackage[usenames,dvipsnames]{xcolor}
% `hyperref` comes after `xr-hyper`.
\usepackage{xr-hyper}