Enderton. Single-rooted/valued proofs.

finite-set-exercises
Joshua Potter 2023-06-24 09:52:45 -06:00
parent 27372eb160
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@ -90,6 +90,8 @@ Given \nameref{ref:relation} $R$, the \textbf{field} of $R$, denoted $\fld{R}$,
A \textbf{function} is a relation $F$ such that for each $x$ in $\dom{F}$ there A \textbf{function} is a relation $F$ such that for each $x$ in $\dom{F}$ there
is only one $y$ such that $xFy$. is only one $y$ such that $xFy$.
In other words, $F$ is \textbf{single-valued}.
We say that $F$ is a function \textbf{from $A$ into $B$} or that $F$ We say that $F$ is a function \textbf{from $A$ into $B$} or that $F$
\textbf{maps $A$ into $B$} (written $F \colon A \rightarrow B$) iff $F$ is a \textbf{maps $A$ into $B$} (written $F \colon A \rightarrow B$) iff $F$ is a
function, $\dom{F} = A$, and $\ran{F} \subseteq B$. function, $\dom{F} = A$, and $\ran{F} \subseteq B$.
@ -3079,7 +3081,7 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
\end{proof} \end{proof}
\subsection{\unverified{Theorem 3F}}% \subsection{\partial{Theorem 3F}}%
\label{sub:theorem-3f} \label{sub:theorem-3f}
\begin{theorem}[3F] \begin{theorem}[3F]
@ -3099,11 +3101,98 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
\lean{Bookshelf/Enderton/Set/Chapter\_3} \lean{Bookshelf/Enderton/Set/Chapter\_3}
{Enderton.Set.Chapter\_3.theorem\_3f\_ii} {Enderton.Set.Chapter\_3.theorem\_3f\_ii}
TODO We prove that (i) any set $F$, $F^{-1}$ is a function iff $F$ is
single-rooted and (ii) any relation $F$ is a function iff $F^{-1}$ is
single-rooted.
\paragraph{(i)}%
\label{par:theorem-3f-i}
Let $F$ be any set.
\subparagraph{($\Rightarrow$)}%
Suppose $F^{-1}$ is a \nameref{ref:function}.
By definition, for each $x \in \dom{(F^{-1})}$, there is only one $y$
such that $\left< x, y \right> \in F^{-1}$.
By definition of the \nameref{ref:inverse} of $F$,
$F^{-1} = \{\left< u, v \right> \mid vFu\}$.
Then for each $x \in \ran{F}$, there exists exactly one $y$ such that
$\left< y, x \right> \in F$.
This definitionally means $F$ is single-rooted.
\subparagraph{($\Leftarrow$)}%
Suppose $F$ is single-rooted.
By definition, for each $x \in \ran{F}$, there is only one $t$ such that
$\left< t, x \right> \in F$.
By definition of the \nameref{ref:inverse} of $F$,
$F^{-1} = \{\left< u, v \right> \mid vFu\}$.
Then for each $x \in \dom{(F^{-1})}$ there exists exactly one $t$ such
that $\left< x, t \right> \in F^{-1}$.
This definitionally means $F^{-1}$ is a function.
\paragraph{(ii)}%
Let $F$ be a \nameref{ref:relation}.
\subparagraph{($\Rightarrow$)}%
Suppose $F$ is a function.
By \nameref{sub:theorem-3e}, $F = (F^{-1})^{-1}$.
Then by \nameref{par:theorem-3f-i}, $F^{-1}$ is single-rooted.
\subparagraph{($\Leftarrow$)}%
Suppose $F^{-1}$ is single-rooted.
Then by \nameref{par:theorem-3f-i}, $(F^{-1})^{-1}$ is a function.
By \nameref{sub:theorem-3e}, $(F^{-1})^{-1} = F$.
Thus $F$ is a function.
\end{proof} \end{proof}
\subsection{\unverified{Theorem 3G}}% \subsection{\partial{Lemma 1}}%
\label{sub:lemma-1}
Assume that $F$ is a one-to-one function.
Then $F^{-1}$ is a one-to-one function.
\begin{proof}
We prove that (i) $F^{-1}$ is a function and (ii) $F^{-1}$ is single-rooted.
\paragraph{(i)}%
\label{par:lemma-1-i}
By hypothesis, $F$ is one-to-one.
This means it is single-rooted, i.e. for all $x \in \ran{F}$, there exists
exactly one $t$ such that $\left< t, x \right> \in F$.
By definition of the \nameref{ref:inverse} of $F$,
$\left< x, t \right> \in F^{-1}$.
But then for all $x \in \dom{(F^{-1})}$, there exists exactly one $t$ such
that $\left< x, t \right> \in F^{-1}$.
Thus $F^{-1}$ is a function.
\paragraph{(ii)}%
\label{par:lemma-1-ii}
By hypothesis, $F$ is single-valued.
That is, for all $x \in \dom{F}$, there exists exactly one $y$ such that
$\left< x, y \right> \in F$.
By definition of the \nameref{ref:inverse} of $F$,
$\left< y, x \right> \in F^{-1}$.
But then for all $x \in \ran{(F^{-1})}$, there exists exactly one $y$ such
that $\left< y, x \right> \in F^{-1}$.
Thus $F^{-1}$ is single-rooted.
\paragraph{Conclusion}%
By \nameref{par:lemma-1-i} and \nameref{par:lemma-1-ii}, $F^{-1}$ is
a one-to-one function.
\end{proof}
\subsection{\partial{Theorem 3G}}%
\label{sub:theorem-3g} \label{sub:theorem-3g}
\begin{theorem}[3G] \begin{theorem}[3G]
@ -3116,7 +3205,17 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
\begin{proof} \begin{proof}
TODO Suppose $F$ is a one-to-one \nameref{ref:function}.
Then \nameref{sub:lemma-1} indicates $F^{-1}$ is a one-to-one function with
domain $\ran{F}$ and range $\dom{F}$.
For all $x \in \dom{F}$, $\left< x, F(x) \right> \in F$.
Then $\left< F(x), x \right> \in F^{-1}$.
Since $F^{-1}$ is single-valued, $F^{-1}(F(x)) = x$ is well-defined.
For all $y \in \ran{F}$, $\left< y, F^{-1}(y) \right> \in F^{-1}$.
Then $\left< F^{-1}(y), y \right> \in F$.
Since $F$ is single-valued, $F(F^{-1}(y)) = y$ is also well-defined.
\end{proof} \end{proof}