Enderton. Add equivalence relation problem prompts.
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@ -32,6 +32,13 @@ For any relation $R$ there is a function $H \subseteq R$ with
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For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
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for all $i \in I$, then $$\bigtimes_{i \in I} H(i) \neq \emptyset.$$
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\section{\pending{Compatible}}%
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\label{sec:compatible}
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A \nameref{ref:function} $F$ is \textbf{compatible} with relation $R$ if and
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only if for all $x$ and $y$ in $A$,
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$$xRy \Rightarrow F(x)RF(y).$$
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\section{\defined{Composition}}%
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\label{ref:composition}
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@ -3269,7 +3276,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{proof}
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\subsection{\sorry{Theorem 3P}}%
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\subsection{\pending{Theorem 3P}}%
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\label{sub:theorem-3p}
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\begin{theorem}[3P]
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@ -3280,6 +3287,50 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\end{theorem}
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\begin{proof}
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Let $\Pi = \{[x]_R \mid x \in A\}$.
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We show that (i) no two different sets in $\Pi$ have any common elements and
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(ii) that each element of $A$ is in some set in $\Pi$.
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\paragraph{(i)}%
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Let $[x]_R, [y]_R \in \Pi$ be two different sets.
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We must show that $[x]_R \cap [y]_R = \emptyset$.
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For the sake of contradiction, suppose $[x]_R \cap [y]_R \neq \emptyset$.
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Let $z \in [x]_R \cap [y]_R$.
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Then $xRz$ and $yRz$.
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Since $R$ is an \nameref{ref:equivalence-relation} on $A$, it is
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\nameref{ref:symmetric} and \nameref{ref:transitive}.
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Then $zRy$ and $xRy$.
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By \nameref{sub:lemma-3n}, $xRy$ if and only if $[x]_R = [y]_R$,
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contradicting the distinctness of $[x]_R$ and $[y]_R$.
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Thus it follows $[x]_R \cap [y]_R] = \emptyset$.
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\paragraph{(ii)}%
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Let $x \in A$.
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Since $R$ is an \nameref{ref:equivalence-relation} on $A$, it follows
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$xRx$.
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Thus $x$ is a member of some set in $\Pi$, namely $[x]_R$.
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\end{proof}
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\subsection{\sorry{Theorem 3Q}}%
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\label{sub:theorem-3q}
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\begin{theorem}[3Q]
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Assume that $R$ is an equivalence relation on $A$ and that
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$F \colon A \rightarrow A$.
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If $F$ is compatible with $R$, then there exists a unique
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$\hat{F} \colon A / R \rightarrow A / R$ such that
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$$\hat{F}([x]_R) = [F(x)]_R \quad\text{for all } x \text{ in } A.$$
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If $F$ is not compatible with $R$, then no such $\hat{F}$ exists.
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Analogous results apply to functions from $A \times A$ into $A$.
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\end{theorem}
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\begin{proof}
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TODO
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@ -4416,4 +4467,164 @@ Show that from the first form of the axiom of choice we can prove the second
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\end{proof}
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\subsection{\sorry{Exercise 3.32}}%
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\label{sub:exercise-3.32}
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\begin{enumerate}[(a)]
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\item Show that $R$ is symmetric iff $R^{-1} \subseteq R$.
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\item Show that $R$ is transitive iff $R \circ R \subseteq R$.
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.33}}%
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\label{sub:exercise-3.33}
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Show that $R$ is a symmetric and transitive relation iff $R = R^{-1} \circ R$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.34}}%
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\label{sub:exercise-3.34}
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Assume that $\mathscr{A}$ is a nonempty set, every member of which is a
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transitive relation.
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\begin{enumerate}[(a)]
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\item Is the set $\bigcap{\mathscr{A}}$ a transitive relation?
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\item Is $\bigcup{\mathscr{A}}$ a transitive relation?
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.35}}%
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\label{sub:exercise-3.35}
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Show that for any $R$ and $x$, we have $[x]_R = \img{R}{\{x\}}$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.36}}%
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\label{sub:exercise-3.36}
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Assume that $f \colon A \rightarrow B$ and that $R$ is an equivalence relation
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on $B$.
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Define $Q$ to be the set
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$$\{\left< x, y \right> \in A \times A \mid
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\left< f(x), f(x) \right> \in R\}.$$
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Show that $Q$ is an equivalence relation on $A$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.37}}%
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\label{sub:exercise-3.37}
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Assume that $\Pi$ is a partition of a set $A$.
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Define the relation $R_\Pi$ as follows:
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$$xR_{\Pi}y \iff (\exists B \in \Pi)(x \in B \land y \in B).$$
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Show that $R_\Pi$ is an equivalence relation on $A$.
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(This is a formalized version of the discussion at the beginning of this
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section.)
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.38}}%
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\label{sub:exercise-3.38}
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Theorem 3P shows that $A / R$ is a partition of $A$ whenever $R$ is an
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equivalence relation on $A$.
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Show that if we start with the equivalence relation $R_\Pi$ of the preceding
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exercise, then the partition $A / R_\Pi$ is just $\Pi$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.39}}%
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\label{sub:exercise-3.39}
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Assume that we start with an equivalence relation $R$ on $A$ and define $\Pi$ to
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be the partition $A / R$.
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Show that $R_\Pi$, as defined in Exercise 37, is just $R$.
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.40}}%
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\label{sub:exercise-3.40}
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Define an equivalence relation $R$ on the set $P$ of positive integers by
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$$mRn \iff m \text{ and } n \text{ have the same number of prime factors}.$$
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Is there a function $f \colon P / R \rightarrow P / R$ such that
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$f([n]_R) = [3n]_R$ for each $n$?
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.41}}%
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\label{sub:exercise-3.41}
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Let $\mathbb{R}$ be the set of real numbers and define the relation $Q$ on
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$\mathbb{R} \times \mathbb{R}$ by $\left< u, v \right>Q \left< x, y \right>$
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iff $u + y = x + v$.
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\begin{enumerate}[(a)]
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\item Show that $Q$ is an equivalence relation on
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$\mathbb{R} \times \mathbb{R}$.
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\item Is there a function $G \colon (\mathbb{R} \times \mathbb{R}) / Q
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\rightarrow (\mathbb{R} \times \mathbb{R}) / Q$ satisfying the equation
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$$G([\left< x, y \right>]_Q) = [\left< x + 2y, y + 2x \right>]_Q?$$
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\end{enumerate}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 3.42}}%
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\label{sub:exercise-3.42}
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State precisely the "analogous results" mentioned in Theorem 3Q.
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(This will require extending the concept of compatibility in a suitable way.)
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\begin{proof}
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TODO
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\end{proof}
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\end{document}
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@ -518,7 +518,8 @@ theorem theorem_3j_a {F : Set.Relation α} {A B : Set α}
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intro y hy
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have ⟨x, hx⟩ := ran_exists hy
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sorry
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· sorry
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· intro h
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sorry
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/-- #### Theorem 3J (b)
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