Enderton. Prove out most of the set function exercises.
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@ -6,11 +6,6 @@
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\input{../../preamble}
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\makeleancommands{../..}
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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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\header{Elements of Set Theory}{Herbert B. Enderton}
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@ -52,8 +47,8 @@ The \textbf{composition} of sets $F$ and $G$ is
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\section{\defined{Domain}}%
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\label{ref:domain}
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Given \nameref{ref:relation} $R$, the \textbf{domain} of $R$, denoted $\dom{R}$,
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is given by $$x \in \dom{R} \iff \exists y \left< x, y \right> \in R.$$
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The \textbf{domain} of set $R$, denoted $\dom{R}$, is given by
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$$x \in \dom{R} \iff \exists y \left< x, y \right> \in R.$$
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\begin{definition}
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@ -77,7 +72,8 @@ There is a set having no members:
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\label{ref:equivalence-relation}
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Relation $R$ is an \textbf{equivalence relation} if and only if $R$ is a binary
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relation that is reflexive, symmetric, and transitive.
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\nameref{ref:relation} that is \nameref{ref:reflexive},
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\nameref{ref:symmetric}, and \nameref{ref:transitive}.
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\section{\defined{Extensionality Axiom}}%
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\label{ref:extensionality-axiom}
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@ -233,8 +229,8 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
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\section{\defined{Range}}%
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\label{ref:range}
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Given \nameref{ref:relation} $R$, the \textbf{range} of $R$, denoted $\ran{R}$,
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is given by $$x \in \ran{R} \iff \exists t \left< t, x \right> \in R.$$
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The \textbf{range} of set $R$, denoted $\ran{R}$, is given by
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$$x \in \ran{R} \iff \exists t \left< t, x \right> \in R.$$
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\begin{definition}
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@ -242,8 +238,8 @@ Given \nameref{ref:relation} $R$, the \textbf{range} of $R$, denoted $\ran{R}$,
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\end{definition}
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\section{\pending{Reflexive Relation}}%
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\label{ref:reflexive-relation}
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\section{\pending{Reflexive}}%
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\label{ref:reflexive}
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A binary relation $R$ is \textbf{reflexive} on $A$ if and only if $xRx$ for all
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$x \in A$.
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@ -284,8 +280,8 @@ For each formula $\phi$ not containing $B$, the following is an axiom:
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\end{axiom}
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\section{\pending{Symmetric Relation}}%
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\label{ref:symmetric-relation}
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\section{\pending{Symmetric}}%
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\label{ref:symmetric}
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A binary relation $R$ is \textbf{symmetric} on $A$ if and only if whenever
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$xRy$ then $yRx$.
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@ -302,8 +298,8 @@ The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set
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\end{definition}
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\section{\pending{Transitive Relation}}%
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\label{ref:transitive-relation}
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\section{\pending{Transitive}}%
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\label{ref:transitive}
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A binary relation $R$ is \textbf{transitive} on $A$ if and only if whenever
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$xRy$ and $yRz$, then $xRz$.
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@ -2885,7 +2881,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\subparagraph{($\Rightarrow$)}%
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Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$.
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All that remains is to prove $F$ is single-valued.
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All that remains is to prove $F$ is single-rooted.
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Let $y \in \ran{F}$.
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By definition of the \nameref{ref:range} of a function, there exists some
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$x$ such that $\left< x, y \right> \in F$.
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@ -3175,7 +3171,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\section{Equivalence Relations}%
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\label{sec:equivalence-relations}
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\subsection{\sorry{Theorem 3M}}%
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\subsection{\pending{Theorem 3M}}%
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\label{sub:theorem-3m}
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\begin{theorem}[3M]
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@ -3187,7 +3183,23 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\begin{proof}
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TODO
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Suppose $R$ is a \nameref{ref:symmetric} and \nameref{ref:transitive}
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\nameref{ref:relation}.
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By definition, the \nameref{ref:field} of $R$ is given by
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$\fld{R} = \dom{R} \cup \ran{R}$.
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An \nameref{ref:equivalence-relation} is, by definition, a
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\nameref{ref:reflexive}, symmetric, and transitive relation.
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Thus all that remains is to show $R$ is reflexive on $\fld{R}$.
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Let $x \in \fld{R}$.
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Then $x \in \dom{R}$ or $x \in \ran{R}$.
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If $x \in \dom{R}$, there exists some $y$ such that $xRy$.
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Since $R$ is symmetric, it follows $yRx$.
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Since $R$ is transitive, it follows $xRx$.
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If instead $x \in \ran{R}$, there exists some $t$ such that $tRx$.
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Since $R$ is symmetric, it follows $xRt$.
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Since $R$ is transitive, it follows $xRx$.
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Thus $R$ is reflexive on $\fld{R}$.
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\end{proof}
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@ -3681,7 +3693,7 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
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\end{answer}
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\subsection{\sorry{Exercise 3.11}}%
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\subsection{\verified{Exercise 3.11}}%
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\label{sub:exercise-3.11}
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Prove the following version (for functions) of the extensionality principle:
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@ -3691,11 +3703,24 @@ Then $F = G$.
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\begin{proof}
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TODO
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\lean{Init/Core}{funext}
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Let $F$ and $G$ be functions such that $\dom{F} = \dom{G}$ and $F(x) = G(x)$
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for all $x$ in the common domain.
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We prove that $\left< x, y \right> \in F$ if and only if
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$\left< x, y \right> \in G$.
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But this follows immediately:
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\begin{align*}
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\left< x, y \right> \in F
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& \iff y = F(x) \land \left< x, F(x) \right> \in F \\
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& \iff y = G(x) \land \left< x, G(x) \right> \in G \\
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& \iff \left< x, y \right> \in G.
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\end{align*}
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By the \nameref{ref:extensionality-axiom}, $F = G$.
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\end{proof}
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\subsection{\sorry{Exercise 3.12}}%
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\subsection{\pending{Exercise 3.12}}%
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\label{sub:exercise-3.12}
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Assume that $f$ and $g$ are functions and show that
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@ -3704,11 +3729,31 @@ Assume that $f$ and $g$ are functions and show that
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\begin{proof}
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TODO
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Let $f$ and $g$ be \nameref{ref:function}s.
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\paragraph{($\Rightarrow$)}%
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Suppose $f \subseteq g$.
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Then for all \nameref{ref:ordered-pair}s $\left< x, y \right>$,
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$\left< x, y \right> \in f$ implies $\left< x, y \right> \in g$.
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Thus every $x \in \dom{f}$ must be a member of $\dom{g}$.
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Likewise, by definition of a function, $f$ and $g$ are single-valued.
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Thus $f(x) = y$ and $g(x) = y$.
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Since $x$ is an arbitrary element in the domain of $f$, it follows
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$(\forall x \in \dom{f}) f(x) = y = g(x)$.
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\paragraph{($\Leftarrow$)}%
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Suppose $\dom{f} \subseteq \dom{g}$ and
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$(\forall x \in \dom{f}) f(x) = g(x)$.
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Let $\left< x, y \right> \in f$.
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By hypothesis, $x \in \dom{g}$ and $y = f(x) = g(x)$.
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Thus $\left< x, y \right> \in g$ as well.
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Therefore $f \subseteq g$.
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\end{proof}
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\subsection{\sorry{Exercise 3.13}}%
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\subsection{\pending{Exercise 3.13}}%
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\label{sub:exercise-3.13}
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Assume that $f$ and $g$ are functions with $f \subseteq g$ and
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@ -3717,11 +3762,17 @@ Show that $f = g$.
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\begin{proof}
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TODO
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Let $f$ and $g$ be functions such that $f \subseteq g$ and
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$\dom{g} \subseteq \dom{f}$.
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By \nameref{sub:exercise-3.12}, it follows that $\dom{f} \subseteq \dom{g}$
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and $(\forall x \in \dom{f}) f(x) = g(x)$.
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Since $\dom{g} \subseteq \dom{f}$ and $\dom{f} \subseteq \dom{g}$, it follows
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that $\dom{g} = \dom{f}$.
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By \nameref{sub:exercise-3.11}, $f = g$.
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\end{proof}
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\subsection{\sorry{Exercise 3.14}}%
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\subsection{\pending{Exercise 3.14}}%
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\label{sub:exercise-3.14}
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Assume that $f$ and $g$ are functions.
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@ -3734,35 +3785,116 @@ Assume that $f$ and $g$ are functions.
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\begin{proof}
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TODO
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Assume $f$ and $g$ are functions.
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\paragraph{(a)}%
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Consider $f \cap g$.
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By definition of the intersection of sets, $f \cap g \subseteq f$.
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By \nameref{sub:exercise-3.12}, $\dom{(f \cap g)} = \dom{f}$ and
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$(\forall x \in \dom{(f \cap g)} (f \cap g)(x) = f(x)$.
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The latter conjunct shows that, since $f$ is single-valued, $f \cap g$ must
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also be single-valued.
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In other words, $f \cap g$ is a function.
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\paragraph{(b)}%
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\subparagraph{($\Rightarrow$)}%
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Suppose $f \cup g$ is a function.
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Let $x \in (\dom{f}) \cap (\dom{g})$.
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That is, $x \in \dom{f}$ and $x \in \dom{g}$.
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Then there exists only one $y_1$ such that $\left< x, y_1 \right> \in f$.
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Likewise there exists only one $y_2$ such that
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$\left< x, y_2 \right> \in g$.
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But $\left< x, y_1 \right> \in f \cup g$ and
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$\left< x, y_2 \right> \in f \cup g$.
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Since $f \cup g$ is single-valued, it follows $y_1 = y_2$.
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That is, $f(x) = g(x)$.
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\subparagraph{($\Leftarrow$)}%
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Suppose $f(x) = g(x)$ for every $x \in (\dom{f}) \cap (\dom{g})$.
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Let $x \in \dom{(f \cup g)}$.
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There are three cases to consider:
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\begin{enumerate}[(i)]
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\item Suppose $x \in \dom{f}$ but not in $\dom{g}$.
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Since $f$ is a function, it follows $f \cup g$ has only one value $y$
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such that $\left< x, y \right> \in f \cup g$.
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\item Suppose $x \in \dom{g}$ but not in $\dom{f}$.
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Again, since $g$ is a function, it follows $f \cup g$ has only one
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value $y$ such that $\left< x, y \right> \in f \cup g$.
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\item Suppose $x \in \dom{f}$ and $x \in \dom{g}$.
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By hypothesis, $f(x) = g(x)$ meaning there is only one value $y$ such
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that $\left< x, y \right> \in f \cup g$.
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\end{enumerate}
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The above cases are exhaustive.
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Together they imply that $f \cup g$ is single-valued, i.e. a function.
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\end{proof}
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\subsection{\sorry{Exercise 3.15}}%
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\subsection{\pending{Exercise 3.15}}%
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\label{sub:exercise-3.15}
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Let $\mathscr{A}$ be a set of functions such that for any $f$ and $g$ in
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$\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$.
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Show that $\bigcup \mathscr{A}$ is a function.
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Show that $\bigcup{\mathscr{A}}$ is a function.
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\begin{proof}
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TODO
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Let $\mathscr{A}$ be a set of \nameref{ref:function}s such that for any $f$
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and $g$ in $\mathscr{A}$, either $f \subseteq g$ or $g \subseteq f$.
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Let $x \in \dom{\bigcup{\mathscr{A}}}$.
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Then there exists some $y_1$ such that
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$\left< x, y_1 \right> \in \bigcup{\mathscr{A}}$.
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Suppose there also exists some $y_2$ such that
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$\left< x, y_2 \right> \in \bigcup{\mathscr{A}}$.
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By definition of the union of sets, there exists some function
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$f \in \mathscr{A}$ such that $\left< x, y_1 \right> \in f$.
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Likewise there exists some function $g \in \mathscr{A}$ such that
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$\left< x, y_2 \right> \in g$.
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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $f \subseteq g$.
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Then $\left< x, y_1 \right>, \left< x, y_2 \right> \in g$.
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Since $g$ is a function, i.e. single-valued, $y_1 = y_2$.
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\paragraph{Case 2}%
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Suppose $g \subseteq f$.
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Then $\left< x, y_1 \right>, \left< x, y_2 \right> \in f$.
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Since $f$ is a function, i.e. single-valued, $y_1 = y_2$.
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\paragraph{Conclusion}%
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Since the above two cases applies for all
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$x \in \dom{\bigcup{\mathscr{A}}}$ and appropriate choices of $f$ and $g$,
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it follows $\bigcup{\mathscr{A}}$ is indeed a function.
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\end{proof}
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\subsection{\sorry{Exercise 3.16}}%
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\subsection{\unverified{Exercise 3.16}}%
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\label{sub:exercise-3.16}
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Show that there is no set to which every function belongs.
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\begin{proof}
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TODO
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Every \nameref{ref:relation} consisting of a single \nameref{ref:ordered-pair}
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is, by definition, a \nameref{ref:function}.
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By \nameref{sub:exercise-3.4}, there is no set to which every ordered pair
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belongs.
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Thus there is no set to which every function of the described type belongs
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either, let alone a set to which \textit{every} function belongs.
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\end{proof}
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\subsection{\sorry{Exercise 3.17}}%
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\subsection{\pending{Exercise 3.17}}%
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\label{sub:exercise-3.17}
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Show that the composition of two single-rooted sets is again single-rooted.
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@ -3770,11 +3902,36 @@ Conclude that the composition of two one-to-one functions is again one-to-one.
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\begin{proof}
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TODO
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Let $F$ and $G$ be two single-rooted sets.
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Consider $F \circ G$.
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By definition of the \nameref{ref:composition} of sets,
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\begin{equation}
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\label{sub:exercise-3.17-eq1}
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F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}.
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\end{equation}
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Consider any $v \in \ran{(F \circ G)}$.
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By definition of the \nameref{ref:range} of a \nameref{ref:relation}, there
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exists some $u_1$ such that $\left< u_1, v \right> \in F \circ G$.
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Let $u_2$ be a set such that $\left< u_2, v \right> \in F \circ G$.
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By \eqref{sub:exercise-3.17-eq1}, there exists a set $t_1$ such that
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$\left< u_1, t_1 \right> \in G$ and $\left< t_1, v \right> \in F$.
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Likewise, there exists a set $t_2$ such that
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$\left< u_2, t_2 \right> \in G$ and $\left< t_2, v \right> \in F$.
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But $F$ is single-rooted, meaning $t_1 = t_2$.
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Likewise, because $G$ is single-rooted, $u_1 = u_2$.
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Thus $F \circ G$ must also be single-rooted.
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\suitdivider
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Let $f$ and $g$ be one-to-one functions.
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By \nameref{sub:theorem-3h}, $f \circ g$ is single-valued.
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By the above, $f \circ g$ is single-rooted.
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Thus $f \circ g$ is one-to-one.
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\end{proof}
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\subsection{\sorry{Exercise 3.18}}%
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\subsection{\pending{Exercise 3.18}}%
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\label{sub:exercise-3.18}
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Let $R$ be the set
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@ -3785,11 +3942,25 @@ Evaluate the following: $R \circ R$, $R \restriction \{1\}$,
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\begin{proof}
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TODO
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\begin{enumerate}[(i)]
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\item $R \circ R = \{
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\left< 0, 2 \right>,
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\left< 0, 3 \right>,
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\left< 0, 3 \right>,
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\left< 1, 3 \right>
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\}$.
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\item $R \restriction \{1\} = \{
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\left< 1, 2 \right>,
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\left< 1, 3 \right>
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\}$.
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\item $R^{-1} \restriction \{1\} = \{\left< 1, 0 \right>\}$.
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\item $\img{R}{\{1\}} = \{2, 3\}$.
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\item $\img{R^{-1}}{\{1\}} = \{0\}$.
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\end{enumerate}
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\end{proof}
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\subsection{\sorry{Exercise 3.19}}%
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\subsection{\pending{Exercise 3.19}}%
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\label{sub:exercise-3.19}
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Let $$A = \{
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@ -3804,33 +3975,80 @@ Evaluate each of the following: $A(\emptyset)$, $\img{A}{\emptyset}$,
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\begin{proof}
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TODO
|
||||
\begin{enumerate}[(i)]
|
||||
\item $A(\emptyset) = \{\emptyset, \{\emptyset\}\}$.
|
||||
\item $\img{A}{\emptyset} = \emptyset$.
|
||||
\item $\img{A}{\{\emptyset\}} = \{\{\emptyset, \{\emptyset\}\}\}$.
|
||||
\item $\img{A}{\{\emptyset, \{\emptyset\}\}} =
|
||||
\{\{\emptyset, \{\emptyset\}\}, \emptyset\}$.
|
||||
\item $A^{-1} = \{
|
||||
\left< \{\emptyset, \{\emptyset\}\}, \emptyset \right>,
|
||||
\left< \emptyset, \{\emptyset\} \right>
|
||||
\}$.
|
||||
\item $A \circ A =
|
||||
\{\left< \{\emptyset\}, \{\emptyset, \{\emptyset\}\} \right>\}$.
|
||||
\item $A \restriction \emptyset = \emptyset$
|
||||
\item $A \restriction \{\emptyset\} =
|
||||
\{\left< \emptyset, \{\emptyset, \{\emptyset\}\} \right>\}$.
|
||||
\item $A \restriction \{\emptyset, \{\emptyset\}\} = A$.
|
||||
\item $\bigcup\bigcup A =
|
||||
\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$.
|
||||
\end{enumerate}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.20}}%
|
||||
\subsection{\pending{Exercise 3.20}}%
|
||||
\label{sub:exercise-3.20}
|
||||
|
||||
Show that $F \restriction A = F \cap (A \times \ran{F})$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $F$ and $A$ be arbitrary sets.
|
||||
By definition of the \nameref{ref:restriction}, intersection,
|
||||
\nameref{ref:range}, and \nameref{sub:cartesian-product} of sets,
|
||||
Then
|
||||
\begin{align*}
|
||||
F \restriction A
|
||||
& = \{\left< u, v \right> \mid uFv \land u \in A\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
uFv \land u \in A \land v \in \ran{F}\} \\
|
||||
& = \{\left< u, v \right> \mid uFv\} \cap
|
||||
\{\left< u, v \right> \mid u \in A \land v \in \ran{F}\} \\
|
||||
& = F \cap \{\left< u, v \right> \mid u \in A \land v \in \ran{F}\} \\
|
||||
& = F \cap (A \times \ran{F}).
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.21}}%
|
||||
\subsection{\pending{Exercise 3.21}}%
|
||||
\label{sub:exercise-3.21}
|
||||
|
||||
Show that $(R \circ S) \circ T = R \circ (S \circ T)$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $R$, $S$, and $T$ be arbitrary sets.
|
||||
By definition of the \nameref{ref:composition} of sets,
|
||||
\begin{align*}
|
||||
(R \circ S) \circ T
|
||||
& = \{\left< u, v \right> \mid
|
||||
\exists t(u(R \circ S)t \land tTv)\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
\exists t((\exists a, uRa \land aSt) \land tTv)\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
\exists t, \exists a, (uRa \land aSt \land tTv)\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
\exists a, \exists t, (uRa \land aSt \land tTv)\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
\exists a(uRa \land (\exists t, aSt \land tTv))\} \\
|
||||
& = \{\left< u, v \right> \mid \exists a(uRa \land a(S \circ T)v)\} \\
|
||||
& = R \circ (S \circ T).
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.22}}%
|
||||
\subsection{\pending{Exercise 3.22}}%
|
||||
\label{sub:exercise-3.22}
|
||||
|
||||
Show that the following are correct for any sets.
|
||||
|
|
@ -3844,11 +4062,55 @@ Show that the following are correct for any sets.
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $A$, $B$, $F$, $G$, and $Q$ be arbitrary sets.
|
||||
|
||||
\paragraph{(a)}%
|
||||
|
||||
Suppose $A \subseteq B$.
|
||||
Let $x \in \img{F}{A}$.
|
||||
By definition of the \nameref{ref:image} of a set,
|
||||
$\img{F}{A} = \{v \mid (\exists u \in A) uFv\}$.
|
||||
Thus there exists some $u \in A$ such that $uFx$.
|
||||
But $A \subseteq B$ meaning $u \in B$.
|
||||
That is, $(\exists u \in B)uFx$.
|
||||
Thus $$x \in \{v \mid (\exists u \in B)uFv\} = \img{F}{B}.$$
|
||||
|
||||
\paragraph{(b)}%
|
||||
|
||||
By definition of the \nameref{ref:composition} and \nameref{ref:image} of a
|
||||
set,
|
||||
\begin{align*}
|
||||
\img{(F \circ G)}{A}
|
||||
& = \{v \mid (\exists u \in A) u(F \circ G)v\} \\
|
||||
& = \{v \mid (\exists u \in A) \left< u, v \right> \in F \circ G\} \\
|
||||
& = \{v \mid (\exists u \in A)
|
||||
\left< u, v \right> \in \{\left< b, c \right> \mid
|
||||
\exists a(bGa \land aFc)\}\} \\
|
||||
& = \{v \mid \exists u \in A, \exists a, uGa \land aFv\} \\
|
||||
& = \{v \mid \exists a, \exists u \in A, uGa \land aFv\} \\
|
||||
& = \{v \mid \exists a, (\exists u \in A, uGa) \land aFv\} \\
|
||||
& = \{v \mid \exists a \in \{w \mid (\exists u \in A)uGw\}, aFv\} \\
|
||||
& = \{v \mid (\exists a \in \img{G}{A}) aFv\} \\
|
||||
& = \img{F}{\img{G}{A}}.
|
||||
\end{align*}
|
||||
|
||||
\paragraph{(c)}%
|
||||
|
||||
By definition of the \nameref{ref:restriction} of a set,
|
||||
\begin{align*}
|
||||
Q \restriction (A \cup B)
|
||||
& = \{\left< u, v \right> \mid uQv \land u \in A \cup B\} \\
|
||||
& = \{\left< u, v \right> \mid uQv \land (u \in A \lor u \in B)\} \\
|
||||
& = \{\left< u, v \right> \mid
|
||||
(uQv \land u \in A) \lor (uQv \land u \in B)\} \\
|
||||
& = \{\left< u, v \right> \mid uQv \land u \in A\} \cup
|
||||
\{\left< u, v \right> \mid uQv \land u \in B\} \\
|
||||
& = (Q \restriction A) \cup (Q \restriction B).
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.23}}%
|
||||
\subsection{\pending{Exercise 3.23}}%
|
||||
\label{sub:exercise-3.23}
|
||||
|
||||
Let $I_A$ be the identity function on the set $A$.
|
||||
|
|
@ -3858,11 +4120,53 @@ Show that for any sets $B$ and $C$,
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $I_A$ be the identity function on the set $A$.
|
||||
That is, $I_A = \{\left< u, u \right> \mid u \in A\}$.
|
||||
Let $B$ and $C$ be any sets.
|
||||
We show that (i) $B \circ I_A = B \restriction A$ and (ii)
|
||||
$\img{I_A}{C} = A \cap C$.
|
||||
|
||||
\paragraph{(i)}%
|
||||
|
||||
We show that $B \circ I_A \subseteq B \restriction A$ and
|
||||
$B \restriction A \subseteq B \circ I_A$.
|
||||
|
||||
\subparagraph{($\subseteq$)}%
|
||||
|
||||
Let $\left< x, y \right> \in B \circ I_A$.
|
||||
By definition of the \nameref{ref:composition} of sets,
|
||||
there exists some $t$ such that $xBt$ and $t(I_A)y$.
|
||||
By definition of the identity function, $I_A(t) = y$ implies $t = y$.
|
||||
Thus $xBy$.
|
||||
By hypothesis, $x \in \dom{(B \circ I_A)}$.
|
||||
Therefore $x \in \dom{I_A} = A$.
|
||||
Thus
|
||||
$$\left< x, y \right>
|
||||
\in \{\left< u, v \right> \mid u \in A \land uBv\}
|
||||
= B \restriction A.$$
|
||||
|
||||
\subparagraph{($\supseteq$)}%
|
||||
|
||||
Let $\left< x, y \right> \in B \restriction A$.
|
||||
By definition of the \nameref{ref:restriction} of sets,
|
||||
$x \in A$ and $xBy$.
|
||||
But $I_A(x) = x$ meaning $\left< I_A(x), y \right> \in B$.
|
||||
In other words, $\left< x, y \right> \in B \circ I_A$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
By definition of the \nameref{ref:image} of sets,
|
||||
\begin{align*}
|
||||
\img{I_A}{C}
|
||||
& = \{v \mid (\exists u \in C) \left< u, v \right> \in I_A\} \\
|
||||
& = \{v \mid \exists u \in C, u \in A \land u = v\} \\
|
||||
& = \{v \mid v \in C \land v \in A\} \\
|
||||
& = C \cap A.
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.24}}%
|
||||
\subsection{\pending{Exercise 3.24}}%
|
||||
\label{sub:exercise-3.24}
|
||||
|
||||
Show that for a function $F$,
|
||||
|
|
@ -3870,11 +4174,20 @@ Show that for a function $F$,
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $F$ be a function.
|
||||
By definition of the \nameref{ref:inverse} of a set,
|
||||
\begin{align*}
|
||||
\img{F^{-1}}{A}
|
||||
& = \{x \mid (\exists y \in A) yF^{-1}x\} \\
|
||||
& = \{x \mid (\exists y \in A) xFy\} \\
|
||||
& = \{x \mid (\exists y \in A) \left< x, y \right> \in F\} \\
|
||||
& = \{x \mid \exists x \in F, F(x) \in A\} \\
|
||||
& = \{x \in \dom{F} \mid F(x) \in A\}.
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.25}}%
|
||||
\subsection{\pending{Exercise 3.25}}%
|
||||
\label{sub:exercise-3.25}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
|
|
@ -3887,22 +4200,38 @@ Show that for a function $F$,
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
\paragraph{(b)}%
|
||||
\label{par:exercise-3.25-b}
|
||||
|
||||
Let $G$ be a function.
|
||||
Let $\left< x, y \right> \in G \circ G^{-1}$.
|
||||
By definition of the \nameref{ref:composition} of sets, there exists some
|
||||
set $t$ such that $x(G^{-1})t$ and $tGy$.
|
||||
By definition of the \nameref{ref:inverse} of a set,
|
||||
$$x(G^{-1})t \iff tGx.$$
|
||||
By hypothesis, $G$ is single-valued.
|
||||
Thus $x = y$.
|
||||
That is, $G \circ G^{-1}$ is the identity function on $\ran{G}$.
|
||||
|
||||
\paragraph{(a)}%
|
||||
|
||||
This immediately follows from part \nameref{par:exercise-3.25-b}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.26}}%
|
||||
\subsection{\pending{Exercise 3.26}}%
|
||||
\label{sub:exercise-3.26}
|
||||
|
||||
Prove the second halves of parts (a) and (b) of Theorem 3K.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Refer to \nameref{sub:theorem-3k-a}, \nameref{sub:theorem-3k-b}, and
|
||||
\nameref{sub:theorem-3k-c}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{\sorry{Exercise 3.27}}%
|
||||
\subsection{\pending{Exercise 3.27}}%
|
||||
\label{sub:exercise-3.27}
|
||||
|
||||
Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$.
|
||||
|
|
@ -3910,7 +4239,36 @@ Show that $\dom{(F \circ G)} = \img{G^{-1}}{\dom{F}}$ for any sets $F$ and $G$.
|
|||
|
||||
\begin{proof}
|
||||
|
||||
TODO
|
||||
Let $F$ and $G$ be arbitrary sets.
|
||||
We show that each side of our desired equality is a subset of the other.
|
||||
|
||||
\paragraph{($\subseteq$)}%
|
||||
|
||||
Let $x \in \dom{(F \circ G)}$.
|
||||
Then there exists a set $y$ such that $\left< x, y \right> \in F \circ G$.
|
||||
By definition of the \nameref{ref:composition} of sets, there exists a set
|
||||
$t$ such that $xGt$ and $tFy$.
|
||||
Thus $t \in \dom{F}$.
|
||||
Therefore
|
||||
\begin{align*}
|
||||
x
|
||||
& \in \{v \mid (\exists t \in \dom{F}) vGt\} \\
|
||||
& = \{v \mid (\exists t \in \dom{F}) t(G^{-1})v\} \\
|
||||
& = \img{G^{-1}}{\dom{F}}.
|
||||
\end{align*}
|
||||
|
||||
\paragraph{($\supseteq$)}%
|
||||
|
||||
Let $x \in \img{G^{-1}}{\dom{F}}$.
|
||||
Then, by definition of the \nameref{ref:image} of a set, there exists some
|
||||
$u \in \dom{F}$ such that $u(G^{-1})x$.
|
||||
By definition of the \nameref{ref:inverse} of a set, $xGu$.
|
||||
By definition of the \nameref{ref:domain} of a set, there exists some $t$
|
||||
such that $uFt$.
|
||||
Thus $xGu \land uFt$.
|
||||
By definition of the \nameref{ref:composition} of sets,
|
||||
$\left< x, t \right> \in F \circ G$.
|
||||
Therefore $x \in \dom{(F \circ G)}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
|
|||
|
|
@ -507,10 +507,18 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
|
|||
**iff** `F` is one-to-one.
|
||||
-/
|
||||
theorem theorem_3j_a {F : Set.Relation α} {A B : Set α}
|
||||
(hF : F.mapsInto A B) (hA : Set.Nonempty A)
|
||||
(hF : F.isSingleValued ∧ F.mapsInto A B) (hA : Set.Nonempty A)
|
||||
: (∃ G : Set.Relation α,
|
||||
G.mapsInto B A ∧ (∀ p ∈ G.comp F, p.1 = p.2)) ↔ F.isOneToOne := by
|
||||
G.isSingleValued ∧ G.mapsInto B A ∧
|
||||
(∀ p ∈ G.comp F, p.1 = p.2)) ↔ F.isOneToOne := by
|
||||
apply Iff.intro
|
||||
· intro ⟨G, ⟨hG₁, hG₂, hI⟩⟩
|
||||
refine ⟨hF.left, ?_⟩
|
||||
show F.isSingleRooted
|
||||
intro y hy
|
||||
have ⟨x, hx⟩ := ran_exists hy
|
||||
sorry
|
||||
· sorry
|
||||
|
||||
/-- #### Theorem 3J (b)
|
||||
|
||||
|
|
@ -519,9 +527,10 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
|
|||
`B` **iff** `F` maps `A` onto `B`.
|
||||
-/
|
||||
theorem theorem_3j_b {F : Set.Relation α} {A B : Set α}
|
||||
(hF : F.mapsInto A B) (hA : Set.Nonempty A)
|
||||
(hF : F.isSingleValued ∧ F.mapsInto A B) (hA : Set.Nonempty A)
|
||||
: (∃ H : Set.Relation α,
|
||||
H.mapsInto B A ∧ (∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by
|
||||
H.isSingleValued ∧ H.mapsInto B A ∧
|
||||
(∀ p ∈ F.comp H, p.1 = p.2)) ↔ F.mapsOnto A B := by
|
||||
sorry
|
||||
|
||||
end
|
||||
|
|
|
|||
|
|
@ -131,12 +131,16 @@
|
|||
|
||||
\newcommand{\abs}[1]{\left|#1\right|}
|
||||
\newcommand{\ceil}[1]{\left\lceil#1\right\rceil}
|
||||
\newcommand{\dom}[1]{\textop{dom}{#1}}
|
||||
\newcommand{\fld}[1]{\textop{fld}{#1}}
|
||||
\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}
|
||||
\newcommand{\icc}[2]{\left[#1, #2\right]}
|
||||
\newcommand{\ico}[2]{\left[#1, #2\right)}
|
||||
\newcommand{\img}[2]{#1\!\left\llbracket#2\right\rrbracket}
|
||||
\newcommand{\ioc}[2]{\left(#1, #2\right]}
|
||||
\newcommand{\ioo}[2]{\left(#1, #2\right)}
|
||||
\newcommand{\powerset}[1]{\mathscr{P}#1}
|
||||
\newcommand{\ran}[1]{\textop{ran}{#1}}
|
||||
\newcommand{\textop}[1]{\mathop{\text{#1}}}
|
||||
\newcommand{\ubar}[1]{\text{\b{$#1$}}}
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue