Enderton. Exercise 3.38.
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@ -112,7 +112,7 @@ There is a set having no members:
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\end{axiom}
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\section{\defined{Equivalence Class}}%
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\label{sec:equivalence-class}
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\label{ref:equivalence-class}
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The set $[x]_R$ is defined by $$[x]_R = \{t \mid xRt\}.$$
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If $R$ is an \nameref{ref:equivalence-relation} and $x \in fld R$, then $[x]_R$
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@ -3409,8 +3409,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\begin{proof}
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\lean{Bookshelf/Enderton/Set/Chapter\_3}
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{Enderton.Set.Chapter\_3.theorem\_3p}
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\lean{Bookshelf/Enderton/Set/Relation}{Set.Relation.modEquiv\_partition}
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Let $\Pi = \{[x]_R \mid x \in A\}$.
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We show that (i) there are no empty sets in $\Pi$, (ii) no two different sets
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@ -5165,7 +5164,7 @@ Show that $R_\Pi$ is an equivalence relation on $A$.
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\end{proof}
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\subsection{\pending{Exercise 3.38}}%
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\subsection{\verified{Exercise 3.38}}%
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\label{sub:exercise-3.38}
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\nameref{sub:theorem-3p} shows that $A / R$ is a partition of $A$ whenever $R$
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@ -5175,26 +5174,90 @@ Show that if we start with the equivalence relation $R_\Pi$ of the preceding
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\begin{proof}
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If $\Pi$ is empty, its trivial to see $A / R_\Pi$ is likewise empty.
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Thus assume $\Pi$ is not empty.
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Let $C \in \Pi$.
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By definition of a \nameref{ref:partition}, $C$ is nonempty.
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Let $x \in C$.
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By definition of a \nameref{ref:partition}, every member of $\Pi$ is disjoint
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from every other.
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Thus
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\begin{align*}
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C
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& = \{t \mid (\exists B \in \Pi)(x \in B \land t \in B)\} \\
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& = \{t \mid x R_\Pi t\} & \textref{sub:exercise-3.37} \\
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& = [x]_{R_\Pi}.
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\end{align*}
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Therefore every member of $\Pi$ is of form $[x]_{R_\Pi}$ for some $x \in A$.
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In other words, $$A / R_\Pi = \{[x]_{R_\Pi} \mid x \in A\} = \Pi.$$
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\lean{Bookshelf/Enderton/Set/Relation}
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{Enderton.Set.Chapter\_3.exercise\_3\_38}
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By definition,
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\begin{equation}
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\label{sub:exercise-3.38-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 $A / R_\Pi = \Pi$.
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By the \nameref{ref:extensionality-axiom}, these two sets are equal when
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$$B \in A / R_\Pi \iff B \in \Pi.$$
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We prove both directions of this biconditional.
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\paragraph{($\Rightarrow$)}%
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Suppose $B \in A / R_\Pi$.
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By \nameref{sub:exercise-3.37}, $R_\Pi$ is an equivalence class.
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Then, by definition of a \nameref{ref:quotient-set},
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$$A / R_\Pi = \{[x]_{R_\Pi} \mid x \in A\},$$
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whose members are the \nameref{ref:equivalence-class}es.
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Thus there exists some $x \in A$ such that $B = [x]_{R_\Pi}$.
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By definition of a \nameref{ref:partition}, there exists a unique set
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$B' \in \Pi$ containing $x$.
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Thus it suffices to prove that $B = B'$, for then $B = [x]_{R_\Pi} = B'$ is
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a member of $\Pi$ as desired.
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We proceed by extensionality again; that is, we show
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$$y \in B \iff y \in B'.$$
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\subparagraph{($\rightarrow$)}%
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Suppose $y \in B$.
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Then
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\begin{align*}
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y
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& \in B = [x]_{R_\Pi} \\
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& = \{t \mid \pair{x, t} \in R_\Pi\} \\
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& = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\}.
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& \eqref{sub:exercise-3.38-eq1}
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\end{align*}
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Thus there exists some $B_1 \in \Pi$ such that $x \in B_1$ and
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$y \in B_1$.
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By definition of a \nameref{ref:partition}, $B_1 \in \Pi$ is the unique
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member of $\Pi$ containing $y$.
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Thus $B_1 = B'$ meaning $y \in B'$ as desired.
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\subparagraph{($\leftarrow$)}%
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Suppose $y \in B'$.
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By construction, $x \in B'$.
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Then there exists a set $B_1$ such that $x \in B_1$ and $y \in B_1$,
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namely $B'$.
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Therefore
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\begin{align*}
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y
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& \in \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\
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& = \{t \mid \pair{x, t} \in R_\Pi\} \\
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& = [x]_{R_\Pi} = B.
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\end{align*}
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\subparagraph{Conclusion}%
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By the \nameref{ref:extensionality-axiom}, it follows $B = B'$.
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Since $B' \in P$, it also follows $B \in P$.
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\paragraph{($\Leftarrow$)}%
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Let $B \in \Pi$.
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By definition of a \nameref{ref:partition}, $B$ is nonempty.
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Let $x \in B$.
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By definition of a set, $B = \{t \mid x \in B \land t \in B\}$.
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By definition of a \nameref{ref:partition}, every member of $B$ must belong
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to only $B$ (i.e. no other sets in the partition).
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Thus we can equivalently write
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\begin{align*}
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B
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& = \{t \mid (\exists B_1 \in \Pi)(x \in B_1 \land t \in B_1)\} \\
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& = \{ t \mid \pair{x, t} \in R_\Pi \} \\
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& = [x]_{R_\Pi}.
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\end{align*}
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Therefore $B \in A / R_{\Pi}$.
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\end{proof}
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\subsection{\pending{Exercise 3.39}}%
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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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@ -5203,22 +5266,7 @@ Show that $R_\Pi$, as defined in Exercise 37, is just $R$.
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\begin{proof}
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If $\Pi$ is empty, it's trivial to show $R_\Pi$ is also empty.
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Thus assume $\Pi$ is not empty.
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Let $C \in \Pi = A / R$.
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By definition of a \nameref{ref:partition}, $C$ is nonempty.
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Let $x \in C$.
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Thus
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\begin{align*}
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C
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& = [x]_R & \textref{sub:theorem-3p} \\
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& = \{t \mid xRt\} \\
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& = \{t \mid (\exists B \in A / R)(x \in B \land t \in B)\} \\
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& = \{t \mid (\exists B \in \Pi)(x \in B \land t \in B)\} \\
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& = \{t \mid x R_\Pi t\}.
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\end{align*}
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Thus $\pair{x, y} \in R_\Pi$ if and only if $\pair{x, y} \in R$.
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By the \nameref{ref:extensionality-axiom}, $R_\Pi = R$.
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TODO
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\end{proof}
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@ -1702,58 +1702,6 @@ theorem theorem_3m {R : Set.Relation α}
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have := hS ht
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exact hT this ht
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/-- #### Theorem 3P
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Assume that `R` is an equivalence relation on `A`. Then the set
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`{[x]_R | x ∈ A}` of all equivalence classes is a partition of `P`.
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-/
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theorem theorem_3p {R : Set.Relation α} {A : Set α} {P : Set (Set α)}
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(hR : R.isEquivalence A) (hP : P = {neighborhood R x | x ∈ A})
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: isPartition P A := by
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refine ⟨?_, ?_, ?_, ?_⟩
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· -- Every member is a subset of `A`.
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intro p hp
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rw [hP] at hp
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simp only [Set.mem_setOf_eq] at hp
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have ⟨x, hx⟩ := hp
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rw [← hx.right]
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show ∀ t, t ∈ neighborhood R x → t ∈ A
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intro t ht
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exact hR.b_on (Or.inr (mem_pair_imp_snd_mem_ran ht))
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· -- Every member is nonempty.
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intro p hp
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rw [hP] at hp
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have ⟨x, hx⟩ := hp
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rw [← hx.right]
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exact ⟨x, hR.refl x hx.left⟩
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· -- Every pair of members is disjoint.
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intro xR hxR yR hyR h
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by_contra nh
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have nh' : Set.Nonempty (xR ∩ yR) := by
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rw [← Set.nmem_singleton_empty]
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exact nh
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have ⟨z, hz⟩ := nh'
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rw [hP] at hxR hyR
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have ⟨x, hx⟩ := hxR
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have ⟨y, hy⟩ := hyR
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rw [← hx.right, ← hy.right] at hz
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unfold neighborhood at hz
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simp only [Set.mem_inter_iff, Set.mem_setOf_eq] at hz
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have hzy : (z, y) ∈ R := hR.symm hz.right
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have hxy : (x, y) ∈ R := hR.trans hz.left hzy
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have := (neighborhood_iff_mem hR hx.left hy.left).mpr hxy
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rw [hx.right, hy.right] at this
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exact absurd this h
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· -- Every element of `A` is in `P`.
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intro x hx
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have := hR.refl x hx
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refine ⟨neighborhood R x, ?_, this⟩
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· rw [hP]
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exact ⟨x, hx, rfl⟩
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/-- #### Exercise 3.32 (a)
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Show that `R` is symmetric **iff** `R⁻¹ ⊆ R`.
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@ -1980,26 +1928,27 @@ 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 `R` as follows:
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Assume that `Π` 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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xRy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
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xRₚy ↔ (∃ B ∈ Π)(x ∈ B ∧ y ∈ B).
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```
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Show that `R` is an equivalence relation on `A`. (This is a formalized version
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Show that `Rₚ` is an equivalence relation on `A`. (This is a formalized version
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of the discussion at the beginning of this section.)
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-/
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theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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(hP : isPartition P A) (R : Set.Relation α)
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(hR : ∀ x y, (x, y) ∈ R ↔ ∃ B ∈ P, x ∈ B ∧ y ∈ B)
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: isEquivalence R A := by
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have hR' : R = { p | ∃ B ∈ P, p.1 ∈ B ∧ p.2 ∈ B } := by
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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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ext p
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have (x, y) := p
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exact hR x y
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refine ⟨?_, ?_, ?_, ?_⟩
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exact hRₚ x y
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· -- `fld R ⊆ A`
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show ∀ x, x ∈ fld R → x ∈ A
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rw [hR']
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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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intro x hx
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unfold fld dom ran at hx
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simp only [
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@ -2018,25 +1967,25 @@ theorem exercise_3_37 {P : Set (Set α)} {A : Set α}
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have := hP.p_subset B hB.left
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exact this hB.right.right
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· -- `isReflexive R A`
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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 [hR]
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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 R`
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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 [hR] 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 [hR]
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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 R`
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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 [hR] 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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@ -2050,10 +1999,78 @@ 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 [hR]
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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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/-- #### 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`. Show that if we start with the equivalence relation
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`Rₚ` of the preceding exercise, then the partition `A / Rₚ` is just `P`.
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-/
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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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ext p
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have (x, y) := p
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exact hRₚ x y
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ext B
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apply Iff.intro
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· intro ⟨x, hx⟩
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have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx.left
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simp only at hB'
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suffices B = B' by
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rw [this]
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exact hB'.left.left
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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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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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exact ⟨B', hB'.left.left, hB'.left.right, hy⟩
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· intro hB
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have ⟨x, hx⟩ := hP.nonempty B hB
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have hx' : x ∈ A := hP.p_subset B hB hx
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refine ⟨x, hx', Eq.symm ?_⟩
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calc B
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_ = {t | x ∈ B ∧ t ∈ B} := by
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ext y
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apply Iff.intro
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· intro hy
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exact ⟨hx, hy⟩
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· intro hy
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exact hy.right
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_ = {t | ∃ B₁ ∈ P, x ∈ B₁ ∧ t ∈ B₁} := by
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ext y
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apply Iff.intro
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· intro hy
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exact ⟨B, hB, hy⟩
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· intro hy
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have ⟨B₁, hB₁⟩ := hy
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have ⟨B', hB'⟩ := partition_mem_mem_eq hP hx'
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simp only [Set.mem_setOf_eq] at *
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have : B = B₁ := by
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have lhs := hB'.right B ⟨hB, hx⟩
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have rhs := hB'.right B₁ ⟨hB₁.left, hB₁.right.left⟩
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rw [lhs, rhs]
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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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simp only [Set.mem_setOf_eq]
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_ = neighborhood Rₚ x := rfl
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end Relation
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end Enderton.Set.Chapter_3
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@ -529,6 +529,14 @@ and the members of the set. It isn't standard in anyway.
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-/
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def neighborhood (R : Relation α) (x : α) := { t | (x, t) ∈ R }
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/--
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The neighborhood with some respect to an equivalence relation `R` on set `A`
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and member `x` contains `x`.
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-/
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theorem neighborhood_self_mem {R : Set.Relation α} {A : Set α} {x : α}
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(hR : isEquivalence R A) (hx : x ∈ A)
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: x ∈ neighborhood R x := hR.refl x hx
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/--
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Assume that `R` is an equivalence relation on `A` and that `x` and `y` belong
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to `A`. Then `[x]_R = [y]_R ↔ xRy`.
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@ -560,6 +568,21 @@ structure isPartition (P : Set (Set α)) (A : Set α) : Prop where
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disjoint : ∀ a ∈ P, ∀ b, b ∈ P → a ≠ b → a ∩ b = ∅
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exhaustive : ∀ a ∈ A, ∃ p, p ∈ P ∧ a ∈ p
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/--
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Membership of sets within `P` is unique.
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-/
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theorem partition_mem_mem_eq {P : Set (Set α)} {A : Set α}
|
||||
(hP : isPartition P A) (hx : x ∈ A)
|
||||
: ∃! B, B ∈ P ∧ x ∈ B := by
|
||||
have ⟨B, hB⟩ := hP.exhaustive x hx
|
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refine ⟨B, hB, ?_⟩
|
||||
intro B₁ hB₁
|
||||
by_contra nB
|
||||
have hB_disj := hP.disjoint B hB.left B₁ hB₁.left (Ne.symm nB)
|
||||
rw [Set.ext_iff] at hB_disj
|
||||
have := (hB_disj x).mp ⟨hB.right, hB₁.right⟩
|
||||
simp at this
|
||||
|
||||
/--
|
||||
The partition `A / R` induced by an equivalence relation `R`.
|
||||
-/
|
||||
|
|
|
|||
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Reference in New Issue