Enderton. Finish remaining theorems 3 set.

finite-set-exercises
Joshua Potter 2023-06-25 09:57:03 -06:00
parent a6e6251627
commit 3db21d7a1a
1 changed files with 113 additions and 8 deletions

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@ -3221,25 +3221,63 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof} \end{proof}
\subsection{\unverified{Theorem 3H}}% \subsection{\partial{Theorem 3H}}%
\label{sub:theorem-3h} \label{sub:theorem-3h}
\begin{theorem}[3H] \begin{theorem}[3H]
Assume that $F$ and $G$ are functions. Assume that $F$ and $G$ are functions.
Then $F \circ G$ is a function, its domain is Then $F \circ G$ is a function, its domain is
$$\{x \in \dom{G} \mid G(x) \in \dom{F}\},$$ and for $x$ in its domain, \begin{equation}
$(F \circ G)(x) = F(G(x))$. \label{sub:theorem-3h-eq1}
\{x \in \dom{G} \mid G(x) \in \dom{F}\},
\end{equation}
and for $x$ in its domain, $(F \circ G)(x) = F(G(x))$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
TODO Let $F$ and $G$ be \nameref{ref:function}s.
By definition of the \nameref{ref:composition} of $F$ and $G$,
\begin{equation}
\label{sub:theorem-3h-eq2}
F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}.
\end{equation}
By construction, $F \circ G$ is a relation.
By the definition of the \nameref{ref:domain} of a relation,
$x \in \dom{(F \circ G)}$ if and only if there exists some $y$ such that
$\left< x, y \right> \in F \circ G$.
We prove that (i) $F \circ G$ is a function with domain satisfying
\eqref{sub:theorem-3h-eq1}, and (ii) $(F \circ G)(x) = F(G(x))$.
\paragraph{(i)}%
\label{par:theorem-3h-i}
By \eqref{sub:theorem-3h-eq2}, there exists some $t$ such that
$\left< x, t \right> \in G$ and $\left< t, y \right> \in F$.
Since $G$ is single-valued, $t$ is uniquely determined by $x$.
Since $F$ is single-valued, $y$ is uniquely determined by $t$.
Therefore, by transitivity, $y$ is uniquely determined by $x$.
Thus $F \circ G$ is single-valued, i.e. $F \circ G$ is a function.
Furthermore, by definition of function application, $t = G(x)$.
Thus
$$\left< x, G(x) \right> \in G \quad\text{and}\quad
\left< G(x), y \right> \in F.$$
This immediately implies \eqref{sub:theorem-3h-eq1} holds true.
\paragraph{(ii)}%
Let $x \in \dom{(F \circ G)}$.
By definition, $\left< x, (F \circ G)(x) \right> \in F \circ G$.
Then \eqref{sub:theorem-3h-eq2} implies $(F \circ G)(x)$ satisfies
$\left< G(x), (F \circ G)(x) \right> \in F$.
This is equivalent to saying $F(G(x)) = (F \circ G)(x)$ as expected.
\end{proof} \end{proof}
\subsection{\unverified{Theorem 3I}}% \subsection{\partial{Theorem 3I}}%
\label{sub:theorem-3i} \label{sub:theorem-3i}
\begin{theorem}[3I] \begin{theorem}[3I]
@ -3250,11 +3288,21 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof} \begin{proof}
TODO By definition of the \nameref{ref:composition} of $F$ and $G$,
$$F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}.$$
By definition of the \nameref{ref:inverse} of a function,
\begin{align*}
(F \circ G)^{-1}
& = \{\left< u, v \right> \mid \exists t (vGt \land tFu)\} \\
& = \{\left< u, v \right> \mid \exists t (tFu \land vGt)\} \\
& = \{\left< u, v \right> \mid
\exists t \left[ u(F^{-1})t \land t(G^{-1})v \right]\} \\
& = G^{-1} \circ F^{-1}.
\end{align*}
\end{proof} \end{proof}
\subsection{\unverified{Theorem 3J}}% \subsection{\partial{Theorem 3J}}%
\label{sub:theorem-3j} \label{sub:theorem-3j}
\begin{theorem}[3J] \begin{theorem}[3J]
@ -3273,7 +3321,64 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\begin{proof} \begin{proof}
TODO Let $F$ be a \nameref{ref:function} from nonempty set $A$ to set $B$.
\paragraph{(a)}%
We prove there exists a function $G \colon B \rightarrow A$ such that
$G \circ F = I_A$ if and only if $F$ is one-to-one.
\subparagraph{($\Rightarrow$)}%
Suppose $G \colon B \rightarrow A$ is a function satisfying
$G \circ F = I_A$.
To prove function $F$ is one-to-one, it is sufficient to prove $F$ is
single-rooted.
Let $y \in \ran{F}$.
Then there there exists some $x_1 \in \dom{F}$ such that $F(x_1) = y$.
Suppose there exists another $x_2 \in \dom{F}$ such that $F(x_2) = y$.
Then $G(F(x_1)) = G(y) = G(F(x_2))$.
But $G \circ F = I_A$ so $x_1 = x_2$.
Thus, for all $y \in \ran{F}$, there exists only one $x$ such that
$\left< x, y \right> \in F$.
That is, $F$ is single-rooted as expected.
\subparagraph{($\Leftarrow$)}%
Suppose $F$ is one-to-one.
Then by \nameref{sub:lemma-1}, $F^{-1} \colon B \rightarrow A$ is also
one-to-one.
We show that $F^{-1} \circ F = I_A$.
Let $x \in \dom{F}$.
By \nameref{sub:theorem-3h}, $(F^{-1} \circ F)(x) = F^{-1}(F(x))$.
By \nameref{sub:theorem-3g}, $F^{-1}(F(x)) = x$.
Thus $F^{-1} \circ F = I_A$.
\paragraph{(b)}%
We prove there exists a function $H \colon B \rightarrow A$ such that
$F \circ H = I_B$ if and only if $F$ maps $A$ onto $B$.
\subparagraph{($\Rightarrow$)}%
Suppose $H \colon B \rightarrow A$ is a function satisfying
$F \circ H = I_B$.
To prove $F$ maps $A$ onto $B$, we must prove $\ran{F} = B$.
But for all $y \in B$, $F(H(y)) = y$.
Thus $y \in \ran{F}$ meaning $B \subseteq \ran{F}$.
Since $F$ maps $A$ into $B$ by hypothesis, $\ran{F} \subseteq B$.
Thus $\ran{F} = B$ as expected.
\subparagraph{($\Leftarrow$)}%
Suppose $F$ maps $A$ onto $B$.
That is, $\ran{F} = B$.
We show that $F^{-1} \colon B \rightarrow A$ satisfies
$F \circ F^{-1} = I_B$.
Let $y \in \ran{F}$.
By \nameref{sub:theorem-3h}, $(F \circ F^{-1})(y) = F(F^{-1}(y))$.
By \nameref{sub:theorem-3g}, $F(F^{-1}(y)) = y$.
Thus $F \circ F^{-1} = I_B$.
\end{proof} \end{proof}