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Update `equinumerous` to infix `equin` command.

pigeonhole-redux
Joshua Potter 2023-09-30 14:29:50 -06:00
parent edef7e9b58
commit c985f9f8a5
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48
Bookshelf/Enderton/Set.tex
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@ -94,9 +94,9 @@
Furthermore, Furthermore,
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item For any sets $A$ and $B$, \item For any sets $A$ and $B$,
$$\card{A} = \card{B} \quad\text{iff}\quad \equinumerous{A}{B}.$$ $$\card{A} = \card{B} \quad\text{iff}\quad A \equin B.$$
\item For a finite set $A$, $\card{A}$ is the \nameref{ref:natural-number} \item For a finite set $A$, $\card{A}$ is the \nameref{ref:natural-number}
$n$ for which $\equinumerous{A}{n}$. $n$ for which $A \equin n$.
\end{enumerate} \end{enumerate}
\lean{Mathlib/Data/Finset/Card} \lean{Mathlib/Data/Finset/Card}
@ -170,9 +170,8 @@
\section{\defined{Equinumerous}}% \section{\defined{Equinumerous}}%
\hyperlabel{ref:equinumerous} \hyperlabel{ref:equinumerous}
A set $A$ is \textbf{equinumerous} to a set $B$ (written A set $A$ is \textbf{equinumerous} to a set $B$ (written $A \equin B$) if and
$\equinumerous{A}{B}$) if and only if there is a one-to-one only if there is a one-to-one \nameref{ref:function} from $A$ onto $B$.
\nameref{ref:function} from $A$ onto $B$.
In other words, there exists a one-to-one correspondence between $A$ and $B$. In other words, there exists a one-to-one correspondence between $A$ and $B$.
\lean*{Mathlib/Init/Function} \lean*{Mathlib/Init/Function}
@ -8797,10 +8796,9 @@
\begin{theorem}[6A] \begin{theorem}[6A]
For any sets $A$, $B$, and $C$, For any sets $A$, $B$, and $C$,
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item $\equinumerous{A}{A}$. \item $A \equin A$.
\item If $\equinumerous{A}{B}$, then $\equinumerous{B}{A}$. \item If $A \equin B$, then $B \equin A$.
\item If $\equinumerous{A}{B}$ and $\equinumerous{B}{C}$, then \item If $A \equin B$ and $B \equin C$, then $A \equin C$.
$\equinumerous{A}{C}$.
\end{enumerate} \end{enumerate}
\end{theorem} \end{theorem}
@ -8826,18 +8824,18 @@
\paragraph{(b)}% \paragraph{(b)}%
Suppose $\equinumerous{A}{B}$. Suppose $A \equin B$.
Then there exists a one-to-one correspondence $F$ between $A$ and $B$. Then there exists a one-to-one correspondence $F$ between $A$ and $B$.
Consider now \nameref{ref:inverse} $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$ Consider now \nameref{ref:inverse} $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$
By \nameref{sub:one-to-one-inverse}, $F^{-1}$ is a one-to-one function. By \nameref{sub:one-to-one-inverse}, $F^{-1}$ is a one-to-one function.
For all $y \in A$, $\tuple{y, F(y)} \in F$. For all $y \in A$, $\tuple{y, F(y)} \in F$.
Then $\tuple{F(y), y} \in F^{-1}$ meaning $F^{-1}$ is onto $A$. Then $\tuple{F(y), y} \in F^{-1}$ meaning $F^{-1}$ is onto $A$.
Hence $F^{-1}$ is a one-to-one correspondence between $B$ and $A$, i.e. Hence $F^{-1}$ is a one-to-one correspondence between $B$ and $A$, i.e.
$\equinumerous{B}{A}$. $B \equin A$.
\paragraph{(c)}% \paragraph{(c)}%
Suppose $\equinumerous{A}{B}$ and $\equinumerous{B}{C}$. Suppose $A \equin B$ and $B \equin C$.
Then there exists a one-to-one correspondence $G$ between $A$ and $B$ and Then there exists a one-to-one correspondence $G$ between $A$ and $B$ and
a one-to-one correspondence $F$ between $B$ and $C$. a one-to-one correspondence $F$ between $B$ and $C$.
By \nameref{sub:one-to-one-composition}, $F \circ G$ is a one-to-one By \nameref{sub:one-to-one-composition}, $F \circ G$ is a one-to-one
@ -8851,7 +8849,7 @@
Then $F(G(x)) = y$. Then $F(G(x)) = y$.
Thus $\ran{(F \circ G)} = C$ meaning $F \circ G$ is onto $C$. Thus $\ran{(F \circ G)} = C$ meaning $F \circ G$ is onto $C$.
Hence $F \circ G$ is a one-to-one correspondence function between $A$ and Hence $F \circ G$ is a one-to-one correspondence function between $A$ and
$C$, i.e. $\equinumerous{A}{C}$. $C$, i.e. $A \equin C$.
\end{proof} \end{proof}
@ -8990,8 +8988,8 @@
$S' \cup T = S$. $S' \cup T = S$.
By definition of a \nameref{ref:finite-set}, $S$ is By definition of a \nameref{ref:finite-set}, $S$ is
\nameref{ref:equinumerous} to a natural number $n$. \nameref{ref:equinumerous} to a natural number $n$.
By \nameref{sub:theorem-6a}, $\equinumerous{S' \cup T}{S}$ which, by the By \nameref{sub:theorem-6a}, $S' \cup T \equin S$ which, by the same
same theorem, implies $\equinumerous{S' \cup T}{n}$. theorem, implies $S' \cup T \equin n$.
Let $f$ be a one-to-one correspondence between $S' \cup T$ and $n$. Let $f$ be a one-to-one correspondence between $S' \cup T$ and $n$.
Then $f \restriction S'$ is a one-to-one correspondence between $S'$ and a Then $f \restriction S'$ is a one-to-one correspondence between $S'$ and a
@ -9088,8 +9086,8 @@
By \nameref{sub:trichotomy-law-natural-numbers}, exactly one of three By \nameref{sub:trichotomy-law-natural-numbers}, exactly one of three
situations is possible: $n = m$, $n < m$, or $m < n$. situations is possible: $n = m$, $n < m$, or $m < n$.
If $n < m$, then $\equinumerous{m}{S}$ and $\equinumerous{S}{n}$. If $n < m$, then $m \equin S$ and $S \equin n$.
By \nameref{sub:theorem-6a}, it follows $\equinumerous{m}{n}$. By \nameref{sub:theorem-6a}, it follows $m \equin n$.
But \nameref{sub:pigeonhole-principle} indicates no natural number is But \nameref{sub:pigeonhole-principle} indicates no natural number is
equinumerous to a proper subset of itself, a contradiction. equinumerous to a proper subset of itself, a contradiction.
If $m < n$, a parallel argument applies. If $m < n$, a parallel argument applies.
@ -9114,7 +9112,7 @@
\begin{equation} \begin{equation}
\hyperlabel{sub:lemma-6f-eq1} \hyperlabel{sub:lemma-6f-eq1}
S = \{n \in \omega \mid \forall C \subset n, S = \{n \in \omega \mid \forall C \subset n,
\exists m < n \text{ such that } \equinumerous{C}{m}\}. \exists m < n \text{ such that } C \equin m\}.
\end{equation} \end{equation}
We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$. We prove that (i) $0 \in S$ and (ii) if $n \in S$ then $n^+ \in S$.
Afterward we prove (iii) the lemma statement. Afterward we prove (iii) the lemma statement.
@ -9171,7 +9169,7 @@
\nameref{ref:inductive-set}. \nameref{ref:inductive-set}.
By \nameref{sub:theorem-4b}, $S = \omega$. By \nameref{sub:theorem-4b}, $S = \omega$.
Therefore, for every proper subset $C$ of a natural number $n$, there Therefore, for every proper subset $C$ of a natural number $n$, there
exists some $m < n$ such that $\equinumerous{C}{n}$. exists some $m < n$ such that $C \equin n$.
\end{proof} \end{proof}
@ -9197,19 +9195,19 @@
some proper subset of $n$. some proper subset of $n$.
By \nameref{sub:lemma-6f}, $\ran{(f \restriction S')}$ is equinumerous to By \nameref{sub:lemma-6f}, $\ran{(f \restriction S')}$ is equinumerous to
some $m < n$. some $m < n$.
Then \nameref{sub:theorem-6a} indicates $\equinumerous{S'}{m}$. Then \nameref{sub:theorem-6a} indicates $S' \equin m$.
Hence $S'$ is a finite set. Hence $S'$ is a finite set.
\end{proof} \end{proof}
\subsection{\sorry{Theorem 6H}}% \subsection{\sorry{Theorem 6H}}%
\hyperlabel{sub:theorem-6h} \hyperlabel{sub:theorem-6h}
Assume that $\equinumerous{K_1}{K_2}$ and $\equinumerous{L_1}{L_2}$. Assume that $K_1 \equin K_2$ and $L_1 \equin L_2$.
\begin{enumerate}[(a)] \begin{enumerate}[(a)]
\item If $K_1 \cap L_1 = K_2 \cap L_2 = \emptyset$, then \item If $K_1 \cap L_1 = K_2 \cap L_2 = \emptyset$, then
$\equinumerous{K_1 \cup L_1}{K_2 \cup L_2}$. $K_1 \cup L_1 \equin K_2 \cup L_2$.
\item $\equinumerous{K_1 \times L_1}{K_2 \times L_2}$. \item $K_1 \times L_1 \equin K_2 \times L_2$.
\item $\equinumerous{^{(L_1)}{K_1}}{^{(L_2)}{K_2}}$. \item $^{(L_1)}{K_1} \equin ^{(L_2)}{K_2}$.
\end{enumerate} \end{enumerate}
\begin{proof} \begin{proof}
@ -9594,7 +9592,7 @@
Suppose $f$ is one-to-one. Suppose $f$ is one-to-one.
Then $f$ is a one-to-one correspondence between $A$ and $\ran{f}$. Then $f$ is a one-to-one correspondence between $A$ and $\ran{f}$.
That is, $\equinumerous{A}{\ran{f}}$. That is, $A \equin \ran{f}$.
Because $f$ maps $A$ onto $A$, $\ran{f} \subseteq A$. Because $f$ maps $A$ onto $A$, $\ran{f} \subseteq A$.
Hence $\ran{f} \subset A$ or $\ran{f} = A$. Hence $\ran{f} \subset A$ or $\ran{f} = A$.
But, by \nameref{sub:corollary-6c}, $\ran{f}$ cannot be a proper subset of But, by \nameref{sub:corollary-6c}, $\ran{f}$ cannot be a proper subset of

2
preamble.tex
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@ -170,7 +170,7 @@
\newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil}
\newcommand{\ctuple}[2]{\left< #1, \cdots, #2 \right>} \newcommand{\ctuple}[2]{\left< #1, \cdots, #2 \right>}
\newcommand{\dom}[1]{\textop{dom}{#1}} \newcommand{\dom}[1]{\textop{dom}{#1}}
\newcommand{\equinumerous}[2]{#1 \approx #2} \newcommand{\equin}{\approx}
\newcommand{\fld}[1]{\textop{fld}{#1}} \newcommand{\fld}[1]{\textop{fld}{#1}}
\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}
\newcommand{\icc}[2]{\left[#1, #2\right]} \newcommand{\icc}[2]{\left[#1, #2\right]}