Enderton. Finish most equivalence class exercises.
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@ -3446,7 +3446,7 @@ 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{\pending{Theorem 3Q}}%
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\subsection{\unverified{Theorem 3Q}}%
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\hyperlabel{sub:theorem-3q}
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\begin{theorem}[3Q]
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@ -3480,7 +3480,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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Since $F$ is compatible, $F(x_1)RF(x_2)$.
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Another application of \nameref{sub:lemma-3n} implies that
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$[F(x_1)]_R = [F(x_2)]_R$.
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Thus $\hat{F}$ is be single-valued.
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Thus $\hat{F}$ is single-valued.
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Uniqueness follows immediately from the \nameref{ref:extensionality-axiom}.
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@ -3491,7 +3491,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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By \nameref{sub:lemma-3n}, $[x]_R = [y]_R$.
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For the sake of contradiction, suppose a function $\hat{F}$ exists satisfying
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\eqref{sub:theorem-3q-eq1}.
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Then $\hat{F}([x]_R) = \hat{F}([y]_R)$ and $[F(x)]_R = [F(y)]_R$.
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Then $\hat{F}([x]_R) = \hat{F}([y]_R)$ meaning $[F(x)]_R = [F(y)]_R$.
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Then \nameref{sub:lemma-3n} implies $F(x)RF(y)$, a contradiction.
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Therefore our original hypothesis must be incorrect.
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That is, there is no function $\hat{F}$ satisfying \eqref{sub:theorem-3q-eq1}.
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@ -4777,10 +4777,22 @@ Assume that $F \colon \powerset{A} \rightarrow \powerset{A}$ and that $F$ has
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Define
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$$B = \bigcap\{X \subseteq A \mid F(X) \subseteq X\} \quad\text{and}\quad
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C = \bigcup\{X \subseteq A \mid X \subseteq F(X)\}.$$
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\begin{enumerate}[(a)]
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\item Show that $F(B) = B$ and $F(C) = C$.
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\item Show that if $F(X) = X$, then $B \subseteq X \subseteq C$.
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\end{enumerate}
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\subsubsection{\sorry{Exercise 3.30 (a)}}%
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\hyperlabel{ssub:exercise-3.30-a}
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Show that $F(B) = B$ and $F(C) = C$.
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\begin{proof}
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TODO
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\end{proof}
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\subsubsection{\sorry{Exercise 3.30 (b)}}%
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\hyperlabel{ssub:exercise-3.30-b}
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Show that if $F(X) = X$, then $B \subseteq X \subseteq C$.
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\begin{proof}
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@ -5257,7 +5269,7 @@ Show that if we start with the equivalence relation $R_\Pi$ of the preceding
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\end{proof}
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\subsection{\sorry{Exercise 3.39}}%
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\subsection{\verified{Exercise 3.39}}%
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\hyperlabel{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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@ -5266,11 +5278,44 @@ 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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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_39}
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By definition,
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\begin{equation}
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\hyperlabel{sub:exercise-3.39-eq1}
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R_\Pi = \{ (x, y) \mid (\exists B \in \Pi)(x \in B \land y \in B) \}.
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\end{equation}
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We prove that $R_\Pi = R$.
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By the \nameref{ref:extensionality-axiom}, these two sets are equal when
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$$(x, y) \in R_\Pi \iff (x, y) \in R.$$
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We prove both directions of this biconditional.
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\paragraph{($\Rightarrow$)}%
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Let $(x, y) \in R_\Pi$.
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By \eqref{sub:exercise-3.39-eq1}, there exists some $B \in \Pi$ such that
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$x \in B$ and $y \in B$.
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Since $\Pi = A / R = \{[x]_R \mid x \in A\}$, there must exist some
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$z \in A$ such that $B = [z]_R$.
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By definition of an \nameref{ref:equivalence-class}, $x \in [z]_R$ implies
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that $zRx$ and $y \in [z]_R$ implies $zRy$.
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Since $R$ is \nameref{ref:symmetric}, $xRz$.
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Since $R$ is \nameref{ref:transitive}, $xRy$.
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\paragraph{($\Leftarrow$)}%
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Let $(x, y) \in R$.
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By definition of an \nameref{ref:equivalence-class}, $x \in [x]_R$ and
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$y \in [x]_R$.
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Note also that $[x]_R \in A / R = \Pi$.
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Thus there exists some $B \in \Pi$ such that $x \in B$ and $y \in B$, namely
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$B = [x]_R$.
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By \eqref{sub:exercise-3.39-eq1}, $(x, y) \in R_\Pi$.
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\end{proof}
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\subsection{\pending{Exercise 3.40}}%
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\subsection{\unverified{Exercise 3.40}}%
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\hyperlabel{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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@ -5294,25 +5339,106 @@ Is there a function $f \colon P / R \rightarrow P / R$ such that
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\end{proof}
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\subsection{\sorry{Exercise 3.41}}%
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\subsection{\unverified{Exercise 3.41}}%
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\hyperlabel{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 $\pair{u, v}Q\pair{x, y}$ iff
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$u + y = x + v$.
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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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$$G([\pair{x, y}]_Q) = [\pair{x + 2y, y + 2x}]_Q?$$
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\subsubsection{\verified{Exercise 3.41a}}%
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\hyperlabel{ssub:exercise-3.41-a}
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\end{enumerate}
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Show that $Q$ is an equivalence relation on $\mathbb{R} \times \mathbb{R}$.
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\begin{proof}
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TODO
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.exercise\_3\_41\_a}
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We show (i) $Q$ is \nameref{ref:reflexive} on $\mathbb{R} \times \mathbb{R}$,
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(ii) $Q$ is \nameref{ref:symmetric}, and (iii) $Q$ is
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\nameref{ref:transitive}.
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\paragraph{(i)}%
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Let $\pair{x, y} \in R \times R$.
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Since $x + y = x + y$, it immediately follows $\pair{x, y}Q\pair{x, y}$.
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Thus $Q$ is reflexive on $\mathbb{R}$.
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\paragraph{(ii)}%
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Let $\pair{\pair{u, v}, \pair{x, y}} \in Q$.
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Then $u + y = x + v$.
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Likewise, $x + v = u + y$.
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This immediately implies that $\pair{\pair{x, y}, \pair{u, v}} \in Q$.
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Thus $Q$ is symmetric.
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\paragraph{(iii)}%
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Let $\pair{\pair{u, v}, \pair{x, y}} \in Q$ and
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$\pair{\pair{x, y}, \pair{a, b}} \in Q$.
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Then $u + y = x + v$ and $x + b = a + y$.
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Rearranging terms, we have $u - v = x - y$ and $x - y = a - b$.
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Thus $u - v = a - b$.
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Rearranging terms once more yields $u + b = a + v$.
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Thus $\pair{\pair{u, v}, \pair{a, b}} \in Q$.
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Therefore $Q$ is transitive.
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\end{proof}
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\subsubsection{\unverified{Exercise 3.41b}}%
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\hyperlabel{ssub:exercise-3.41-b}
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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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\begin{equation}
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\hyperlabel{ssub:exercise-3.41-b-eq1}
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G([\pair{x, y}]_Q) = [\pair{x + 2y, y + 2x}]_Q?
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\end{equation}
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\begin{proof}
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Let $f \colon \mathbb{R} \times \mathbb{R}
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\rightarrow \mathbb{R} \times \mathbb{R}$ be given by
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$f(\pair{x, y}) = \pair{x + 2y, y + 2x}$.
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We show (i) that $f$ is \nameref{ref:compatible} with $Q$ and then (ii)
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there exists such a function satisfying \eqref{ssub:exercise-3.41-b-eq1}.
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\paragraph{(i)}%
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\hyperlabel{par:exercise-3.41-b-i}
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Let $\pair{u, v}, \pair{x, y} \in \mathbb{R} \times \mathbb{R}$ such that
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$\pair{u, v} Q \pair{x, y}$.
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Thus
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\begin{equation}
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\hyperlabel{ssub:exercise-3.41-b-eq2}
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u + y = x + v
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\end{equation}
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Next consider
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\begin{align*}
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f(\pair{u, v}) & = \pair{u + 2v, v + 2u}, \\
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f(\pair{x, y}) & = \pair{x + 2y, y + 2x}.
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\end{align*}
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Then
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\begin{align*}
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u + y & = x + v \\
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\iff 3u + 3y & = 3x + 3v \\
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\iff (u + y) + (2u + 2y) & = (x + v) + (2x + 2v) \\
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\iff (x + v) + (2u + 2y) & = (u + y) + (2x + 2v)
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& \eqref{ssub:exercise-3.41-b-eq2} \\
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\iff (x + 2y) + (v + 2u) & = (u + 2v) + (y + 2x) \\
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\iff (u + 2v) + (y + 2x) & = (x + 2y) + (v + 2u).
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\end{align*}
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This last equality shows $f(\pair{u, v}) \,Q\, f(\pair{x, y})$.
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Thus $f$ is compatible with $Q$.
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\paragraph{(ii)}%
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By \nameref{par:exercise-3.41-b-i} and \nameref{sub:theorem-3q}, there
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exists a unique $G \colon (\mathbb{R} \times \mathbb{R}) / Q \rightarrow
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(\mathbb{R} \times \mathbb{R}) / Q$ satisfying
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\eqref{ssub:exercise-3.41-b-eq1}.
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\end{proof}
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@ -2,6 +2,7 @@ import Bookshelf.Enderton.Set.Chapter_2
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import Bookshelf.Enderton.Set.OrderedPair
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import Bookshelf.Enderton.Set.Relation
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import Common.Logic.Basic
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import Mathlib.Data.Real.Basic
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import Mathlib.Tactic.CasesM
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/-! # Enderton.Set.Chapter_3
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@ -1928,7 +1929,7 @@ theorem exercise_3_36 {f : Set.HRelation α β}
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/-- #### Exercise 3.37
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Assume that `Π` is a partition of a set `A`. Define the relation `Rₚ` as
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Assume that `P` is a partition of a set `A`. Define the relation `Rₚ` as
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follows:
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```
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xRₚy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
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@ -1940,7 +1941,7 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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(hP : isPartition P A) (Rₚ : Set.Relation α)
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(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
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: isEquivalence Rₚ A := by
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have hR : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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ext p
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have (x, y) := p
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exact hRₚ x y
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@ -1948,7 +1949,7 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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refine ⟨?_, ?_, ?_, ?_⟩
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· -- `fld Rₚ ⊆ A`
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show ∀ x, x ∈ fld Rₚ → x ∈ A
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rw [hR]
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rw [hRₚ_eq]
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intro x hx
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unfold fld dom ran at hx
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simp only [
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@ -1969,23 +1970,23 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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· -- `isReflexive Rₚ A`
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intro x hx
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rw [hR]
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rw [hRₚ_eq]
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simp only [Set.mem_setOf_eq, and_self]
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exact hP.exhaustive x hx
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· -- `isSymmetric Rₚ`
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intro x y h
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rw [hR] at h
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rw [hRₚ_eq] at h
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simp only [Set.mem_setOf_eq] at h
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have ⟨B, hB⟩ := h
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rw [hR]
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rw [hRₚ_eq]
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simp only [Set.mem_setOf_eq]
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conv at hB => right; rw [and_comm]
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exact ⟨B, hB⟩
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· -- `isTransitive Rₚ`
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intro x y z hx hy
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rw [hR] at hx hy
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rw [hRₚ_eq] at hx hy
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simp only [Set.mem_setOf_eq] at hx hy
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have ⟨B₁, hB₁⟩ := hx
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have ⟨B₂, hB₂⟩ := hy
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@ -1999,7 +2000,7 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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intro h'
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rw [Set.ext_iff] at h'
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exact (h' y).mp ⟨hy₁, hy₂⟩
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rw [hR]
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rw [hRₚ_eq]
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simp only [Set.mem_setOf_eq]
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exact ⟨B₁, hB₁.left, hB₁.right.left, by rw [hB]; exact hB₂.right.right⟩
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@ -2013,7 +2014,7 @@ theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
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(hP : isPartition P A) (Rₚ : Set.Relation α)
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(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
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: modEquiv (exercise_3_37 hP Rₚ hRₚ) = P := by
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have hR : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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ext p
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have (x, y) := p
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exact hRₚ x y
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@ -2030,13 +2031,13 @@ theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
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ext y
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apply Iff.intro
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· intro hy
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rw [← hx.right, hR] at hy
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rw [← hx.right, hRₚ_eq] at hy
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have ⟨B₁, hB₁⟩ := hy
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have := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
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rw [← this]
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exact hB₁.right.right
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· intro hy
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rw [← hx.right, hR]
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rw [← hx.right, hRₚ_eq]
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exact ⟨B', hB'.left.left, hB'.left.right, hy⟩
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· intro hB
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@ -2067,10 +2068,101 @@ theorem exercise_3_38 {P : Set (Set α)} {A : Set α}
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rw [this]
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exact hB₁.right
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_ = {t | (x, t) ∈ Rₚ } := by
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rw [hR]
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rw [hRₚ_eq]
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simp only [Set.mem_setOf_eq]
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_ = neighborhood Rₚ x := rfl
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/-- #### Exercise 3.39
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Assume that we start with an equivalence relation `R` on `A` and define `P` to
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be the partition `A / R`. Show that `Rₚ`, as defined in Exercise 37, is just
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`R`.
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-/
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theorem exercise_3_39 {P : Set (Set α)} {R Rₚ : Set.Relation α} {A : Set α}
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(hR : isEquivalence R A) (hP : P = modEquiv hR)
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(hRₚ : ∀ x y, (x, y) ∈ Rₚ ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
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: Rₚ = R := by
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have hRₚ_eq : Rₚ = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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ext p
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have (x, y) := p
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exact hRₚ x y
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ext p
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have (x, y) := p
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apply Iff.intro
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· intro hp
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rw [hRₚ_eq] at hp
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have ⟨B, hB, hx, hy⟩ := hp
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rw [hP] at hB
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have ⟨z, hz⟩ := hB
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rw [← hz.right] at hx hy
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exact neighborhood_mem_imp_relate hR hx hy
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· intro hp
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have hxA : x ∈ A := hR.b_on (Or.inl (mem_pair_imp_fst_mem_dom hp))
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have hyA : y ∈ A := hR.b_on (Or.inr (mem_pair_imp_snd_mem_ran hp))
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rw [hRₚ_eq]
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have hx : x ∈ neighborhood R x := neighborhood_self_mem hR hxA
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have hy : y ∈ neighborhood R x := by
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rw [← neighborhood_eq_iff_mem_relate hR hxA hyA] at hp
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rw [hp]
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exact neighborhood_self_mem hR hyA
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refine ⟨neighborhood R x, ?_, ⟨hx, hy⟩⟩
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rw [hP]
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exact ⟨x, hxA, rfl⟩
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lemma test {x y z : ℝ} (h : x + y = z) : (x = z - y) := by apply?
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/-- #### Exercise 3.41 (a)
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Let `ℝ` be the set of real numbers and define the realtion `Q` on `ℝ × ℝ` by
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`⟨u, v⟩ Q ⟨x, y⟩` **iff** `u + y = x + v`. Show that `Q` is an equivalence
|
||||
relation on `ℝ × ℝ`.
|
||||
-/
|
||||
theorem exercise_3_41_a {Q : Set.Relation (ℝ × ℝ)}
|
||||
(hQ : ∀ u v x y, ((u, v), (x, y)) ∈ Q ↔ u + y = x + v)
|
||||
: isEquivalence Q (Set.univ : Set (ℝ × ℝ)) := by
|
||||
have hQ_eq : Q = {p | p.1.1 + p.2.2 = p.2.1 + p.1.2} := by
|
||||
ext p
|
||||
apply Iff.intro <;>
|
||||
· intro hp
|
||||
rwa [hQ] at *
|
||||
|
||||
refine ⟨?_, ?_, ?_, ?_⟩
|
||||
|
||||
· -- `fld Q ⊆ Set.univ`
|
||||
show ∀ p, p ∈ fld Q → p ∈ Set.univ
|
||||
intro _ _
|
||||
simp only [Set.mem_univ]
|
||||
|
||||
· -- `isReflexive Q Set.univ`
|
||||
intro (x, y) _
|
||||
rw [hQ_eq]
|
||||
simp
|
||||
|
||||
· -- `isSymmetric Q`
|
||||
intro (u, v) (x, y) hp
|
||||
rw [hQ_eq] at *
|
||||
exact Eq.symm hp
|
||||
|
||||
· -- `isTransitive Q`
|
||||
unfold isTransitive
|
||||
intro (u, v) (x, y) (a, b) h₁ h₂
|
||||
rw [hQ_eq] at *
|
||||
simp at h₁ h₂
|
||||
simp
|
||||
have h₁' : u - v = x - y := by
|
||||
have := sub_eq_of_eq_add h₁
|
||||
rw [add_sub_right_comm] at this
|
||||
exact eq_sub_of_add_eq this
|
||||
have h₂' : x - y = a - b := by
|
||||
have := sub_eq_of_eq_add h₂
|
||||
rw [add_sub_right_comm] at this
|
||||
exact eq_sub_of_add_eq this
|
||||
rw [h₂'] at h₁'
|
||||
have := eq_add_of_sub_eq' h₁'
|
||||
rw [← add_sub_assoc] at this
|
||||
have := add_eq_of_eq_sub this
|
||||
conv => right; rw [add_comm]
|
||||
exact this
|
||||
|
||||
end Relation
|
||||
|
||||
end Enderton.Set.Chapter_3
|
||||
|
|
@ -533,7 +533,7 @@ def neighborhood (R : Relation α) (x : α) := { t | (x, t) ∈ R }
|
|||
The neighborhood with some respect to an equivalence relation `R` on set `A`
|
||||
and member `x` contains `x`.
|
||||
-/
|
||||
theorem neighborhood_self_mem {R : Set.Relation α} {A : Set α} {x : α}
|
||||
theorem neighborhood_self_mem {R : Set.Relation α} {A : Set α}
|
||||
(hR : isEquivalence R A) (hx : x ∈ A)
|
||||
: x ∈ neighborhood R x := hR.refl x hx
|
||||
|
||||
|
|
@ -541,7 +541,7 @@ theorem neighborhood_self_mem {R : Set.Relation α} {A : Set α} {x : α}
|
|||
Assume that `R` is an equivalence relation on `A` and that `x` and `y` belong
|
||||
to `A`. Then `[x]_R = [y]_R ↔ xRy`.
|
||||
-/
|
||||
theorem neighborhood_iff_mem {R : Set.Relation α} {A : Set α} {x y : α}
|
||||
theorem neighborhood_eq_iff_mem_relate {R : Set.Relation α} {A : Set α}
|
||||
(hR : isEquivalence R A) (_ : x ∈ A) (hy : y ∈ A)
|
||||
: neighborhood R x = neighborhood R y ↔ (x, y) ∈ R := by
|
||||
apply Iff.intro
|
||||
|
|
@ -558,6 +558,19 @@ theorem neighborhood_iff_mem {R : Set.Relation α} {A : Set α} {x y : α}
|
|||
· intro ht
|
||||
exact hR.trans h ht
|
||||
|
||||
/--
|
||||
Assume that `R` is an equivalence relation on `A`. If two sets `x` and `y`
|
||||
belong to the same neighborhood, then `xRy`.
|
||||
-/
|
||||
theorem neighborhood_mem_imp_relate {R : Set.Relation α} {A : Set α}
|
||||
(hR : isEquivalence R A)
|
||||
(hx : x ∈ neighborhood R z) (hy : y ∈ neighborhood R z)
|
||||
: (x, y) ∈ R := by
|
||||
unfold neighborhood at hx hy
|
||||
simp only [mem_setOf_eq] at hx hy
|
||||
have := hR.symm hx
|
||||
exact hR.trans this hy
|
||||
|
||||
/--
|
||||
A **partition** `Π` of a set `A` is a set of nonempty subsets of `A` that is
|
||||
disjoint and exhaustive.
|
||||
|
|
@ -622,8 +635,8 @@ theorem modEquiv_partition {A : Set α} {R : Relation α} (hR : isEquivalence R
|
|||
have : z ∈ fld R := Or.inr (mem_pair_imp_snd_mem_ran hz.left)
|
||||
exact hR.b_on this
|
||||
rw [
|
||||
← neighborhood_iff_mem hR hx.left hz_mem,
|
||||
← neighborhood_iff_mem hR hy.left hz_mem,
|
||||
← neighborhood_eq_iff_mem_relate hR hx.left hz_mem,
|
||||
← neighborhood_eq_iff_mem_relate hR hy.left hz_mem,
|
||||
hx.right, hy.right
|
||||
] at hz
|
||||
rw [hz.left, hz.right] at nXY
|
||||
|
|
|
|||
Loading…
Reference in New Issue