Finite set exercises.
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A \textbf{binary operation} on a set $A$ is a \nameref{ref:function} from
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$A \times A$ into $A$.
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\section{\defined{Cardinal Arithmetic}}%
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\hyperlabel{sec:cardinal-arithmetic}
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Let $\kappa$ and $\lambda$ be any cardinal numbers.
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\begin{enumerate}[(a)]
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\item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any
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disjoint sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are
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any sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa^\lambda = \card{(^L{K})}$, where $K$ and $L$ are any sets of
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cardinality $\kappa$ and $\lambda$, respectively.
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\end{enumerate}
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\lean{Mathlib/SetTheory/Cardinal/Basic}
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{Cardinal.add\_def}
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\lean{Mathlib/SetTheory/Cardinal/Basic}
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{Cardinal.mul\_def}
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\lean{Mathlib/SetTheory/Cardinal/Basic}
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{Cardinal.power\_def}
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\section{\defined{Cardinal Number}}%
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\hyperlabel{ref:cardinal-number}
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For any set $C$, the \textbf{cardinal number} of set $C$ is denoted as
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$\card{C}$.
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Furthermore,
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\begin{enumerate}[(a)]
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\item For any sets $A$ and $B$,
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$$\card{A} = \card{B} \quad\text{iff}\quad \equinumerous{A}{B}.$$
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\item For a finite set $A$, $\card{A}$ is the \nameref{ref:natural-number}
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$n$ for which $\equinumerous{A}{n}$.
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\end{enumerate}
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\lean{Mathlib/Data/Finset/Card}
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{Finset.card}
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\lean{Mathlib/SetTheory/Cardinal/Basic}
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{Cardinal}
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\section{\defined{Cartesian Product}}%
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\hyperlabel{ref:cartesian-product}
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\lean{Mathlib/Data/Set/Prod}{Set.prod}
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\section{\defined{Cardinal Arithmetic}}%
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\hyperlabel{sec:cardinal-arithmetic}
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Let $\kappa$ and $\lambda$ be any cardinal numbers.
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\begin{enumerate}[(a)]
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\item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any
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disjoint sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are
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any sets of cardinality $\kappa$ and $\lambda$, respectively.
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\item $\kappa^\lambda = \card{^L{K}}$, where $K$ and $L$ are any sets of
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cardinality $\kappa$ and $\lambda$, respectively.
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\end{enumerate}
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\section{\defined{Compatible}}%
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\hyperlabel{ref:compatible}
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We proceed by contradiction.
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Suppose there existed a set $A$ consisting of every singleton.
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Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set.
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But this set is precisely the class of all sets, which is \textit{not} a
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But this "set" is precisely the class of all sets, which is \textit{not} a
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set.
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Thus our original assumption was incorrect.
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That is, there is no set to which every singleton belongs.
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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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\subsection{\unverified{Exercise 6.6}}%
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\hyperlabel{sub:exercise-6.6}
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Let $\kappa$ be a nonzero cardinal number.
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belongs.
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\begin{proof}
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TODO
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Let $\kappa$ be a nonzero cardinal number and define
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$$\mathbf{K}_\kappa = \{ X \mid \card{X} = \kappa \}.$$
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For the sake of contradiction, suppose $\mathbf{K}_\kappa$ is a set.
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Then the \nameref{ref:union-axiom} suggests $\bigcup \mathbf{K}_{\kappa}$ is
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a set.
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But this "set" is precisely the class of all sets, which is \textit{not} a
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set.
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Thus our original assumption was incorrect.
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That is, there does not exist a set to which every set of cardinality
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$\kappa$ belongs.
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\end{proof}
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\subsection{\sorry{Exercise 6.7}}%
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\subsection{\pending{Exercise 6.7}}%
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\hyperlabel{sub:exercise-6.7}
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Assume that $A$ is finite and $f \colon A \rightarrow A$.
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Show that $f$ is one-to-one iff $\ran{f} = A$.
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\begin{proof}
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TODO
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Let $A$ be a \nameref{ref:finite-set} and $f \colon A \rightarrow A$.
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\paragraph{($\Rightarrow$)}%
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Suppose $f$ is one-to-one.
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Then $f$ is a one-to-one correspondence between $A$ and $\ran{f}$.
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That is, $\equinumerous{A}{\ran{f}}$.
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Because $f$ maps $A$ onto $A$, $\ran{f} \subseteq A$.
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Hence $\ran{f} \subset A$ or $\ran{f} = A$.
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But, by \nameref{sub:corollary-6c}, $\ran{f}$ cannot be a proper subset of
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$A$.
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Thus $\ran{f} = A$.
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\paragraph{($\Leftarrow$)}%
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Suppose $\ran{f} = A$.
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.8}}%
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\subsection{\pending{Exercise 6.8}}%
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\hyperlabel{sub:exercise-6.8}
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Prove that the union of two finite sets is finite, without any use of
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TODO
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\end{proof}
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\subsection{\sorry{Exercise 6.9}}%
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\subsection{\pending{Exercise 6.9}}%
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\hyperlabel{sub:exercise-6.9}
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Prove that the Cartesian product of two finite sets is finite, without any use
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