Enderton. Additional equivalence relation definitions.

finite-set-exercises
Joshua Potter 2023-07-01 15:35:22 -06:00
parent 8294d9583c
commit 48ed3cd59e
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@ -71,8 +71,8 @@ There is a set having no members:
\section{\pending{Equivalence Relation}}%
\label{ref:equivalence-relation}
Relation $R$ is an \textbf{equivalence relation} if and only if $R$ is a binary
\nameref{ref:relation} that is \nameref{ref:reflexive},
Relation $R$ is an \textbf{equivalence relation} on set $A$ if and only if
$R$ is a binary \nameref{ref:relation} that is \nameref{ref:reflexive} on $A$,
\nameref{ref:symmetric}, and \nameref{ref:transitive}.
\section{\defined{Extensionality Axiom}}%
@ -202,6 +202,16 @@ For any sets $u$ and $v$, there is a set having as members just $u$ and $v$:
\end{axiom}
\section{\pending{Partition}}%
\label{ref:partition}
A \textbf{partition} $\Pi$ of a set $A$ is a set of nonempty subsets of $A$ that
is disjoint and exhaustive, i.e.
\begin{enumerate}[(a)]
\item no two different sets in $\Pi$ have any common elements, and
\item each element of $A$ is in some set in $\Pi$.
\end{enumerate}
\section{\defined{Power Set}}%
\label{ref:power-set}
@ -226,6 +236,14 @@ For any set $a$, there is a set whose members are exactly the subsets of $a$:
\end{axiom}
\section{\pending{Quotient Set}}%
\label{ref:quotient-set}
If $R$ is an \nameref{ref:equivalence-relation} on set $A$, then we can define
the \textbf{quotient set} $$A / R = \{[x]_R \mid x \in A\}$$ whose members are
the equivalence classes.
The expression $A / R$ is read "$A$ modulo $R$.
\section{\defined{Range}}%
\label{ref:range}
@ -283,8 +301,8 @@ For each formula $\phi$ not containing $B$, the following is an axiom:
\section{\pending{Symmetric}}%
\label{ref:symmetric}
A binary relation $R$ is \textbf{symmetric} on $A$ if and only if whenever
$xRy$ then $yRx$.
A binary relation $R$ is \textbf{symmetric} if and only if whenever $xRy$ then
$yRx$.
\section{\defined{Symmetric Difference}}%
\label{ref:symmetric-difference}
@ -301,15 +319,16 @@ The \textbf{symmetric difference} $A + B$ of sets $A$ and $B$ is the set
\section{\pending{Transitive}}%
\label{ref:transitive}
A binary relation $R$ is \textbf{transitive} on $A$ if and only if whenever
$xRy$ and $yRz$, then $xRz$.
A binary relation $R$ is \textbf{transitive} if and only if whenever $xRy$ and
$yRz$, then $xRz$.
\section{\defined{Union Axiom}}%
\label{ref:union-axiom}
For any set $A$, there exists a set $B$ whose elements are exactly the members
of the members of $A$:
$$\forall A, \exists B, \forall x \left[ x \in B \iff (\exists b \in A) x \in b \right]$$
$$\forall A, \exists B, \forall x
\left[ x \in B \iff (\exists b \in A) x \in b \right]$$
\begin{axiom}
@ -3187,9 +3206,10 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\nameref{ref:relation}.
By definition, the \nameref{ref:field} of $R$ is given by
$\fld{R} = \dom{R} \cup \ran{R}$.
An \nameref{ref:equivalence-relation} is, by definition, a
\nameref{ref:reflexive}, symmetric, and transitive relation.
Thus all that remains is to show $R$ is reflexive on $\fld{R}$.
An \nameref{ref:equivalence-relation} on $\fld{R}$ is, by definition, a
binary relation \nameref{ref:reflexive} on $\fld{R}$, symmetric, and
transitive.
All that remains is to show $R$ is reflexive on $\fld{R}$.
Let $x \in \fld{R}$.
Then $x \in \dom{R}$ or $x \in \ran{R}$.
@ -3203,6 +3223,69 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
\end{proof}
\subsection{\pending{Lemma 3N}}%
\label{sub:lemma-3n}
\begin{lemma}[3N]
Assume that $R$ is an equivalence relation on $A$ and that $x$ and $y$ belong
to $A$.
Then $$[x]_R = [y]_R \iff xRy.$$
\end{lemma}
\begin{proof}
Suppose $R$ is an \nameref{ref:equivalence-relation} on set $A$.
Let $x, y \in A$.
\paragraph{($\Rightarrow$)}%
Suppose $[x]_R = [y]_R$.
Since $R$ is an equivalence relation, it is reflexive on $A$.
Thus $yRy$ meaning $y \in [y]_R = \{t \mid yRt\}$.
Since $[x]_R = [y]_R$, $y \in \{t \mid xRt\}$ as well.
That is, $xRy$.
\paragraph{($\Leftarrow$)}%
Suppose $xRy$.
We show $[x]_R \subseteq [y]_R$ and $[y]_R \subseteq [x]_R$.
\subparagraph{($\subseteq$)}%
Let $t \in [x]_R$.
Then $xRt$.
Since $R$ is symmetric, $xRy$ implies $yRx$.
Since $R$ is transitive, $yRx$ and $xRt$ implies $yRt$.
Thus $t \in [y]_R$.
\subparagraph{($\supseteq$)}%
Let $t \in [y]_R$.
Then $yRt$.
Since $R$ is transitive, $xRy$ and $yRt$ implies $xRt$.
Thus $t \in [x]_R$.
\end{proof}
\subsection{\sorry{Theorem 3P}}%
\label{sub:theorem-3p}
\begin{theorem}[3P]
Assume that $R$ is an equivalence relation on $A$.
Then the set $\{[x]_R \mid x \in A\}$ of all equivalence classes is a
partition of $A$.
\end{theorem}
\begin{proof}
TODO
\end{proof}
\section{Exercises 3}%
\label{sec:exercises-3}