Enderton (set). Equinumerosity theorems.
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@ -127,8 +127,10 @@
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\section{\defined{Equinumerous}}%
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\hyperlabel{ref:equinumerous}
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A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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if and only if there is a one-to-one \nameref{ref:function} from $A$ onto $B$.
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A set $A$ is \textbf{equinumerous} to a set $B$ (written
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$\equinumerous{A}{B}$) if and only if there is a one-to-one
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\nameref{ref:function} from $A$ onto $B$.
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In other words, there exists a one-to-one correspondence between $A$ and $B$.
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\lean*{Mathlib/Init/Function}
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{Function.Bijective}
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@ -149,8 +151,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\hyperlabel{ref:equivalence-relation}
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Relation $R$ is an \textbf{equivalence relation} on set $A$ if and only if
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$R$ is a binary \nameref{ref:relation} on $A$ that is \nameref{ref:reflexive}
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on $A$, \nameref{ref:symmetric}, and \nameref{ref:transitive}.
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$R$ is a binary \nameref{ref:relation} on $A$ that is
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\nameref{ref:reflexive} on $A$, \nameref{ref:symmetric}, and
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\nameref{ref:transitive}.
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\code*{Bookshelf/Enderton/Set/Relation}
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{Set.Relation.isEquivalence}
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@ -215,6 +218,9 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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only one $x$ such that $xFy$.
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One-to-one functions are sometimes called \textbf{injections}.
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A one-to-one function from $A$ onto $B$ is a
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\textbf{one-to-one correspondence} between $A$ and $B$.
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\code*{Bookshelf/Enderton/Set/Relation}
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{Set.Relation.isSingleValued}
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@ -2879,7 +2885,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\hyperlabel{sub:one-to-one-inverse}
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\begin{lemma}
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For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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For any one-to-one function $F$, $F^{-1}$ is also a one-to-one function.
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\end{lemma}
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\code{Bookshelf/Enderton/Set/Relation}
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@ -3008,7 +3014,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\begin{lemma}
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Let $F$ and $G$ be one-to-one functions.
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Then $F \circ G$ is one-to-one as well.
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Then $F \circ G$ is also a one-to-one function.
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\end{lemma}
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\code{Bookshelf/Enderton/Set/Relation}
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@ -3023,21 +3029,15 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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F \circ G = \{\tuple{u, v} \mid \exists t(uGt \land tFv)\}.
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\end{equation}
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By \nameref{sub:theorem-3h}, $F \circ G$ is a function.
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All that remains is proving $F \circ G$ is single-rooted, i.e. for each
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$y \in \ran{F \circ G}$, there is only one $x$ such that
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$\tuple{x, y} \in F \circ G$.
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To that end let $y \in \ran{(F \circ G)}$.
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Then there exists some $x_1 \in \dom{(F \circ G)}$ such that
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$\tuple{x_1, y} \in F \circ G$.
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Suppose there also exists some $x_2 \in \dom{(F \circ G)}$ such that
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$\tuple{x_2, y} \in F \circ G$.
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By \eqref{sub:one-to-one-composition-eq1}, there exists a $t_1$ such that
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All that remains is proving $F \circ G$ is one-to-one.
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Let $(F \circ G)(x_1) = (F \circ G)(x_2) = y$.
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By \eqref{sub:one-to-one-composition-eq1}, there exists some $t_1$ such that
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$x_1Gt_1$ and $t_1Fy$.
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Likewise, there exists some $t_2$ such that $x_2Gt_2$ and $t_2Fy$.
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But $F$ is one-to-one meaning $t_1 = t_2$.
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Similarly, $G$ is one-to-one meaning $x_1 = x_2$.
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Hence $F \circ G$ is single-valued.
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Since $F$ is one-to-one, it follows $t_1 = t_2$.
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Then, since $G$ is also one-to-one, it follows $x_1 = x_2$.
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Hence $F \circ G$ is one-to-one.
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\end{proof}
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\subsection{\verified{Theorem 3I}}%
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@ -8442,7 +8442,7 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\section{Equinumerosity}%
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\hyperlabel{sec:equinumerosity}
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\subsection{\pending{Theorem 6A}}%
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\subsection{\verified{Theorem 6A}}%
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\hyperlabel{sub:theorem-6a}
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\begin{theorem}[6A]
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@ -8470,23 +8470,43 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\paragraph{(a)}%
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Consider function $I_A \colon A \rightarrow A$ given by $I_A(x) = x$.
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This function is trivially a bijection between $A$ and $A$.
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Consider \nameref{ref:function} $I_A \colon A \rightarrow A$ given by
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$I_A(x) = x$.
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$I_A$ is trivially a one-to-one correspondence between $A$ and $A$.
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Thus $A$ is \nameref{ref:equinumerous} to $A$.
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\paragraph{(b)}%
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Suppose $\equinumerous{A}{B}$.
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Then there exists a one-to-one function $f$ from $A$ onto $B$.
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TODO
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Then there exists a one-to-one correspondence $F$ between $A$ and $B$.
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Consider now \nameref{ref:inverse} $$F^{-1} = \{\tuple{u, v} \mid vFu\}.$$
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By \nameref{sub:one-to-one-inverse}, $F^{-1}$ is a one-to-one function.
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For all $y \in A$, $\tuple{y, F(y)} \in F$.
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Then $\tuple{F(y), y} \in F^{-1}$ meaning $F^{-1}$ is onto $A$.
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Hence $F^{-1}$ is a one-to-one correspondence between $B$ and $A$, i.e.
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$\equinumerous{B}{A}$.
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\paragraph{(c)}%
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TODO
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Suppose $\equinumerous{A}{B}$ and $\equinumerous{B}{C}$.
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Then there exists a one-to-one correspondence $G$ between $A$ and $B$ and
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a one-to-one correspondence $F$ between $B$ and $C$.
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By \nameref{sub:one-to-one-composition}, $F \circ G$ is a one-to-one
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function.
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Thus we're left with proving $F \circ G$ is onto $C$.
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Let $y \in C$.
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Since $F$ is onto $C$, there exists some $t \in B$ such that $F(t) = y$.
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Likewise, since $G$ is onto $B$, there exists some $x \in A$ such that
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$G(x) = t$.
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Then $F(G(x)) = y$.
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Thus $\ran{(F \circ G)} = C$ meaning $F \circ G$ is onto $C$.
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Hence $F \circ G$ is a one-to-one correspondence function between $A$ and
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$C$, i.e. $\equinumerous{A}{C}$.
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\end{proof}
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\subsection{\sorry{Theorem 6B}}%
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\subsection{\unverified{Theorem 6B}}%
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\hyperlabel{sub:theorem-6b}
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\begin{theorem}[6B]
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@ -8494,7 +8514,16 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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\end{theorem}
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\begin{proof}
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TODO
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Let $A$ be an arbitrary set and $f \colon A \rightarrow \powerset{A}$.
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Define $\phi = \{a \in A \mid a \not\in f(a)\}$.
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Clearly $\phi \in \powerset{A}$.
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Furthermore, for all $a \in A$, $\phi \neq f(a)$ since $a \in \phi$ if and
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only if $a \not\in f(a)$.
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Thus $f$ cannot be onto $\powerset{A}$.
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Since $f$ was arbitrarily chosen, there exists no one-to-one correspondence
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between $A$ and $\powerset{A}$.
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Since $A$ was arbitrarily chosen, there is no set equinumerous to its
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powerset.
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\end{proof}
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\section{Finite Sets}%
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@ -8619,13 +8648,13 @@ A set $A$ is \textbf{equinumerous} to a set $B$ (written $\equinumerous{A}{B}$)
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.5}}%
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\subsection{\verified{Exercise 6.5}}%
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\hyperlabel{sub:exercise-6-5}
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Prove \nameref{sub:theorem-6a}.
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\begin{proof}
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TODO
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Refer to \nameref{sub:theorem-6a}.
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\end{proof}
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\subsection{\sorry{Exercise 6.6}}%
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@ -18,22 +18,22 @@ For any sets `A`, `B`, and `C`,
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-/
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theorem theorem_6a_a (A : Set α)
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: ∃ f, Set.BijOn f A A := by
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: ∃ F, Set.BijOn F A A := by
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refine ⟨fun x => x, ?_⟩
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unfold Set.BijOn Set.MapsTo Set.InjOn Set.SurjOn
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simp only [imp_self, implies_true, Set.image_id', true_and]
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exact Eq.subset rfl
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theorem theorem_6a_b [Nonempty α] (A : Set α) (B : Set β)
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(f : α → β) (hf : Set.BijOn f A B)
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: ∃ g, Set.BijOn g B A := by
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refine ⟨Function.invFunOn f A, ?_⟩
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exact (Set.bijOn_comm $ Set.BijOn.invOn_invFunOn hf).mpr hf
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(F : α → β) (hF : Set.BijOn F A B)
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: ∃ G, Set.BijOn G B A := by
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refine ⟨Function.invFunOn F A, ?_⟩
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exact (Set.bijOn_comm $ Set.BijOn.invOn_invFunOn hF).mpr hF
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theorem theorem_6a_c (A : Set α) (B : Set β) (C : Set γ)
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(f : α → β) (hf : Set.BijOn f A B)
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(g : β → γ) (hg : Set.BijOn g B C)
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: ∃ h, Set.BijOn h A C := by
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exact ⟨g ∘ f, Set.BijOn.comp hg hf⟩
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(F : α → β) (hF : Set.BijOn F A B)
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(G : β → γ) (hG : Set.BijOn G B C)
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: ∃ H, Set.BijOn H A C := by
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exact ⟨G ∘ F, Set.BijOn.comp hG hF⟩
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end Enderton.Set.Chapter_6
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