Finite set exercises.

finite-set-exercises
Joshua Potter 2023-09-20 13:39:59 -06:00
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A \textbf{binary operation} on a set $A$ is a \nameref{ref:function} from A \textbf{binary operation} on a set $A$ is a \nameref{ref:function} from
$A \times A$ into $A$. $A \times A$ into $A$.
\section{\defined{Cardinal Arithmetic}}%
\hyperlabel{sec:cardinal-arithmetic}
Let $\kappa$ and $\lambda$ be any cardinal numbers.
\begin{enumerate}[(a)]
\item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any
disjoint sets of cardinality $\kappa$ and $\lambda$, respectively.
\item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are
any sets of cardinality $\kappa$ and $\lambda$, respectively.
\item $\kappa^\lambda = \card{(^L{K})}$, where $K$ and $L$ are any sets of
cardinality $\kappa$ and $\lambda$, respectively.
\end{enumerate}
\lean{Mathlib/SetTheory/Cardinal/Basic}
{Cardinal.add\_def}
\lean{Mathlib/SetTheory/Cardinal/Basic}
{Cardinal.mul\_def}
\lean{Mathlib/SetTheory/Cardinal/Basic}
{Cardinal.power\_def}
\section{\defined{Cardinal Number}}%
\hyperlabel{ref:cardinal-number}
For any set $C$, the \textbf{cardinal number} of set $C$ is denoted as
$\card{C}$.
Furthermore,
\begin{enumerate}[(a)]
\item For any sets $A$ and $B$,
$$\card{A} = \card{B} \quad\text{iff}\quad \equinumerous{A}{B}.$$
\item For a finite set $A$, $\card{A}$ is the \nameref{ref:natural-number}
$n$ for which $\equinumerous{A}{n}$.
\end{enumerate}
\lean{Mathlib/Data/Finset/Card}
{Finset.card}
\lean{Mathlib/SetTheory/Cardinal/Basic}
{Cardinal}
\section{\defined{Cartesian Product}}% \section{\defined{Cartesian Product}}%
\hyperlabel{ref:cartesian-product} \hyperlabel{ref:cartesian-product}
@ -77,19 +118,6 @@
\lean{Mathlib/Data/Set/Prod}{Set.prod} \lean{Mathlib/Data/Set/Prod}{Set.prod}
\section{\defined{Cardinal Arithmetic}}%
\hyperlabel{sec:cardinal-arithmetic}
Let $\kappa$ and $\lambda$ be any cardinal numbers.
\begin{enumerate}[(a)]
\item $\kappa + \lambda = \card{(K \cup L)}$, where $K$ and $L$ are any
disjoint sets of cardinality $\kappa$ and $\lambda$, respectively.
\item $\kappa \cdot \lambda = \card{(K \times L)}$, where $K$ and $L$ are
any sets of cardinality $\kappa$ and $\lambda$, respectively.
\item $\kappa^\lambda = \card{^L{K}}$, where $K$ and $L$ are any sets of
cardinality $\kappa$ and $\lambda$, respectively.
\end{enumerate}
\section{\defined{Compatible}}% \section{\defined{Compatible}}%
\hyperlabel{ref:compatible} \hyperlabel{ref:compatible}
@ -1876,7 +1904,7 @@
We proceed by contradiction. We proceed by contradiction.
Suppose there existed a set $A$ consisting of every singleton. Suppose there existed a set $A$ consisting of every singleton.
Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set. Then the \nameref{ref:union-axiom} suggests $\bigcup A$ is a set.
But this set is precisely the class of all sets, which is \textit{not} a But this "set" is precisely the class of all sets, which is \textit{not} a
set. set.
Thus our original assumption was incorrect. Thus our original assumption was incorrect.
That is, there is no set to which every singleton belongs. That is, there is no set to which every singleton belongs.
@ -9530,7 +9558,7 @@
Refer to \nameref{sub:theorem-6a}. Refer to \nameref{sub:theorem-6a}.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 6.6}}% \subsection{\unverified{Exercise 6.6}}%
\hyperlabel{sub:exercise-6.6} \hyperlabel{sub:exercise-6.6}
Let $\kappa$ be a nonzero cardinal number. Let $\kappa$ be a nonzero cardinal number.
@ -9538,20 +9566,46 @@
belongs. belongs.
\begin{proof} \begin{proof}
TODO Let $\kappa$ be a nonzero cardinal number and define
$$\mathbf{K}_\kappa = \{ X \mid \card{X} = \kappa \}.$$
For the sake of contradiction, suppose $\mathbf{K}_\kappa$ is a set.
Then the \nameref{ref:union-axiom} suggests $\bigcup \mathbf{K}_{\kappa}$ is
a set.
But this "set" is precisely the class of all sets, which is \textit{not} a
set.
Thus our original assumption was incorrect.
That is, there does not exist a set to which every set of cardinality
$\kappa$ belongs.
\end{proof} \end{proof}
\subsection{\sorry{Exercise 6.7}}% \subsection{\pending{Exercise 6.7}}%
\hyperlabel{sub:exercise-6.7} \hyperlabel{sub:exercise-6.7}
Assume that $A$ is finite and $f \colon A \rightarrow A$. Assume that $A$ is finite and $f \colon A \rightarrow A$.
Show that $f$ is one-to-one iff $\ran{f} = A$. Show that $f$ is one-to-one iff $\ran{f} = A$.
\begin{proof} \begin{proof}
Let $A$ be a \nameref{ref:finite-set} and $f \colon A \rightarrow A$.
\paragraph{($\Rightarrow$)}%
Suppose $f$ is one-to-one.
Then $f$ is a one-to-one correspondence between $A$ and $\ran{f}$.
That is, $\equinumerous{A}{\ran{f}}$.
Because $f$ maps $A$ onto $A$, $\ran{f} \subseteq A$.
Hence $\ran{f} \subset A$ or $\ran{f} = A$.
But, by \nameref{sub:corollary-6c}, $\ran{f}$ cannot be a proper subset of
$A$.
Thus $\ran{f} = A$.
\paragraph{($\Leftarrow$)}%
Suppose $\ran{f} = A$.
TODO TODO
\end{proof} \end{proof}
\subsection{\sorry{Exercise 6.8}}% \subsection{\pending{Exercise 6.8}}%
\hyperlabel{sub:exercise-6.8} \hyperlabel{sub:exercise-6.8}
Prove that the union of two finite sets is finite, without any use of Prove that the union of two finite sets is finite, without any use of
@ -9561,7 +9615,7 @@
TODO TODO
\end{proof} \end{proof}
\subsection{\sorry{Exercise 6.9}}% \subsection{\pending{Exercise 6.9}}%
\hyperlabel{sub:exercise-6.9} \hyperlabel{sub:exercise-6.9}
Prove that the Cartesian product of two finite sets is finite, without any use Prove that the Cartesian product of two finite sets is finite, without any use