Enderton. Definitions, theorem 3E.
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\chapter{Reference}%
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\label{chap:reference}
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\section{\partial{Composition}}%
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\label{ref:composition}
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The \textbf{composition} of sets $F$ and $G$ is
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$$F \circ G = \{\left< u, v \right> \mid \exists t(uGt \land tFv)\}.$$
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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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\begin{definition}
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\lean*{Common/Set/Relation}{Set.Relation.dom}
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\end{definition}
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\section{\defined{Empty Set Axiom}}%
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\label{ref:empty-set-axiom}
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@ -49,6 +67,18 @@ If two sets have exactly the same members, then they are equal:
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\end{axiom}
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\section{\defined{Field}}%
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\label{ref:field}
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Given \nameref{ref:relation} $R$, the \textbf{field} of $R$, denoted $\fld{R}$,
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is given by $$\fld{R} = \dom{R} \cup \ran{R}.$$
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\begin{definition}
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\lean*{Common/Set/Relation}{Set.Relation.fld}
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\end{definition}
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\section{\defined{Function}}%
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\label{ref:function}
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@ -75,6 +105,23 @@ One-to-one functions are sometimes called \textbf{injections}.
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\end{definition}
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\section{\partial{Image}}%
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\label{ref:image}
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Let $A$ and $F$ be arbitrary sets.
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The \textbf{image of $A$ under $F$} is the set
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\begin{align*}
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F[A]
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& = \ran{(F \restriction A)} \\
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& = \{v \mid (\exists u \in A) uFv\}.
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\end{align*}
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\section{\partial{Inverse}}%
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\label{ref:inverse}
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The \textbf{inverse} of a set $F$ is the set
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$$F^{-1} = \{\left< u, v \right> \mid vFu\}.$$
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\section{\defined{Ordered Pair}}%
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\label{ref:ordered-pair}
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@ -144,30 +191,35 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
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\end{axiom}
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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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\begin{definition}
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\lean*{Common/Set/Relation}{Set.Relation.ran}
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\end{definition}
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\section{\defined{Relation}}%
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\label{ref:relation}
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A \textbf{relation} is a set of \nameref{ref:ordered-pair}s.
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Given relation $R$, the \textbf{domain} of $R$ ($\dom{R}$), the \textbf{range}
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of $R$ ($\ran{R}$), and the \textbf{field} of $R$ ($\fld{R}$) is given by
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\begin{align*}
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x \in \dom{R} & \iff \exists y \left< x, y \right> \in R, \\
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x \in \ran{R} & \iff \exists t \left< t, x \right> \in R, \\
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\fld{R} & = \dom{R} \cup \ran{R}.
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\end{align*}
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\begin{definition}
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\statementpadding
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\lean*{Mathlib/Data/Rel}{Rel}
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\lean*{Mathlib/Data/Rel}{Rel.dom}
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\lean*{Mathlib/Data/Rel}{Rel.codom}
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\lean*{Common/Set/Relation}{Set.Relation}
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\end{definition}
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\section{\partial{Restriction}}%
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\label{ref:restriction}
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The \textbf{restriction} of a set $F$ to set $A$ is the set
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$$F \restriction A = \{\left< u, v \right> \mid uFv \land u \in A\}.$$
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\section{\defined{Subset Axioms}}%
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\label{ref:subset-axioms}
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@ -2730,8 +2782,8 @@ Show that a set $A$ is a relation iff $A \subseteq \dom{A} \times \ran{A}$.
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Let $a \in A$.
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Since $A$ is a relation, $a$ is an ordered pair.
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Then there exists some sets $x$ and $y$ such that $a = \left< x, y \right>$.
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By definition of the domain and range of $A$, $x \in \dom{A}$ and
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$y \in \ran{A}$.
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By the definition of the \nameref{ref:domain} and \nameref{ref:range} of
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$A$, $x \in \dom{A}$ and $y \in \ran{A}$.
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Thus $a = \left< x, y \right> \in \dom{A} \times \ran{A}$ as well.
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This proves $A \subseteq \dom{A} \times \ran{A}$.
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@ -2885,14 +2937,14 @@ Discuss the result of replacing the union operation by the intersection
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\paragraph{(i)}%
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Let $x \in \dom{\bigcap{\mathscr{A}}}$.
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By definition of the domain of a set,
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By definition of the \nameref{ref:domain} of a set,
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$\exists y, \left< x, y \right> \in \bigcap{\mathscr{A}}$.
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By definition of the intersection of sets,
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$\exists y, \forall R \in \mathscr{A}, \left< x, y \right> \in R$.
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But this implies that
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$\forall R \in \mathscr{A}, \exists y, \left< x, y \right> \in R$.
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By another application of the definition of the domain of a set,
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$\forall R \in \mathscr{A}, x \in \dom{R}$.
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By another application of the definition of the \nameref{ref:domain} of a
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set, $\forall R \in \mathscr{A}, x \in \dom{R}$.
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By another application of the intersection of sets,
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$x \in \bigcap\;\{ \dom{R} \mid R \in \mathscr{A} \}$.
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Thus \eqref{sub:exercise-6.9-eq1} holds.
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@ -2900,14 +2952,14 @@ Discuss the result of replacing the union operation by the intersection
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\paragraph{(ii)}%
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Let $x \in \ran{\bigcap{\mathscr{A}}}$.
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By definition of the range of a set,
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By definition of the \nameref{ref:range} of a set,
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$\exists t, \left< t, x \right> \in \bigcap{\mathscr{A}}$.
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By definition of the intersection of sets,
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$\exists t, \forall R \in \mathscr{A}, \left< t, x \right> \in R$.
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But this implies that
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$\forall R \in \mathscr{A}, \exists t, \left< t, x \right> \in R$.
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By another application of the definition of the domain of a set,
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$\forall R \in \mathscr{A}, x \in \ran{R}$.
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By another application of the definition of the \nameref{ref:range} of a
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set, $\forall R \in \mathscr{A}, x \in \ran{R}$.
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By another application of the intersection of sets,
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$x \in \bigcap\;\{ \ran{R} \mid R \in \mathscr{A} \}$.
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Thus \eqref{sub:exercise-6.9-eq2} holds.
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@ -2945,19 +2997,50 @@ Show that an ordered $4$-tuple is also an ordered $m$-tuple for every positive
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\section{Functions}%
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\label{sec:functions}
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\subsection{\unverified{Theorem 3E}}%
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\subsection{\partial{Theorem 3E}}%
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\label{sub:theorem-3e}
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\begin{theorem}[3E]
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For a set $F$, $\dom{F^{-1}} = \ran{F}$ and $\ran{F^{-1}} = \dom{F}$.
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For a set $F$, $\dom{(F^{-1})} = \ran{F}$ and $\ran{(F^{-1})} = \dom{F}$.
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For a relation $F$, $(F^{-1})^{-1} = F$.
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\end{theorem}
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\begin{proof}
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TODO
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We prove that (i) $\dom{(F^{-1})} = \ran{F}$, (ii) $\ran{(F^{-1})} = \dom{F}$,
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and (iii) $(F^{-1})^{-1} = F$.
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\paragraph{(i)}%
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By definition of the \nameref{ref:domain}, $x \in \dom{(F^{-1})}$ if and
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only if there exists some $y$ such that $\left< x, y \right> \in F^{-1}$.
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By definition of the \nameref{ref:inverse} of a set,
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$\left< y, x \right> \in F$.
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By definition of the \nameref{ref:range}, $x \in \ran{F}$.
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Since each step holds biconditionally, it follows
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$\dom{(F^{-1})} = \ran{F}$ as expected.
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\paragraph{(ii)}%
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By definition of the \nameref{ref:range}, $x \in \ran{(F^{-1})}$ if and
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only if there exists some $t$ such that $\left< t, x \right> \in F^{-1}$.
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By definition of the \nameref{ref:inverse} of a set,
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$\left< x, t \right> \in F$.
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By definition of the \nameref{ref:domain}, $x \in \dom{F}$.
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Since each step holds biconditionally, it follows
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$\ran{(F^{-1})} = \dom{F}$.
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\paragraph{(iii)}%
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By definition of the \nameref{ref:inverse} of a set,
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\begin{align*}
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(F^{-1})^{-1}
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& = \{\left< u, v \right> \mid \left< v, u \right> \in F^{-1}\} \\
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& = \{\left< u, v \right> \mid \left< u, v \right> \in F\} \\
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& = F.
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\end{align*}
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\end{proof}
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