Let $S=\{a,b,c,d\}$ and let $\mathscr{P}=\{ \{ a \}, \{ b,c \}, \{ d \} \}$. Note that $\mathscr{P}$ is a partition of $S$. Describe the equivlanece relation $\textbf{R}$ on $S$ determined by $\mathscr{P}$ as indicated in Theorem 6.15.
Theorem 6.15 states, "Let $\textbf{R}$ be an equivalence relation on a set $S$. Then $\{ E_x:x \in S \}$ is a partition of $S$. The relation 'belongs to the same piece as' is the same as $\textbf{R}$. Converseley, if $\mathscr{P}$ is a partition of $S$, let $\textbf{R}$ be defined by $x\textbf{R}y$ iff $x$ and $y$ are in the same piece of the partition. Then $\textbf{R}$ is an equivalence relation and the corresponding partition into equivalence classes is the same as $\mathscr{P}$"
A relation acts on a set if $(x,y)\in \textbf{R}$. Since $\mathscr{P}$ has three members that must mean there are three pieces of the partition (and thus three seperate equivalence classes). $E_h=\{y \in S:y\textbf{R}h \}$, $E_i=\{y\in S: y\textbf{R}i \}$, $E_j=\{y\in S: y \textbf{R} j \}$.
Each piece of the partition (i think) can be used to construct the ordered pairs of the members of the relation $\textbf{R}$ by simply using the members of each equivalence class in all possible combinations of ordered pairs similar to below.
$\mathscr{P}=\{ \underbrace{\{ a \}}_{(a,a)}, \underbrace{\{ b,c \}}_{(b,b),(b,c),(c,b),(cc)}, \underbrace{\{ d \}}_{(d,d)} \}$
This yields the relation to be $\textbf{R}=\{(a,a),(b,b),(b,c),(c,b),(c,c),(d,d) \}$ which according to the book is the correct answer. However I pretty much worked this out going backwards (starting from the answer and working my way to the original practice problem) so I am unsure if my understanding of the difference between pieces of a partition and thee members of the equivalence class are correct. Is there a more formal way to figure out the equivalence relation $\textbf{R}$?
Best Answer
The equivalence relation $\mathbf{R}$ is defined by
This is easily seen to be an equivalence relation that induces the same partition on $S$.
Thus $$ \mathbf{R}=\bigcup_{E\in\mathscr{P}}(E\times E) $$