Fundamental Theorem on Equivalence Relations/Examples

From ProofWiki
Jump to navigation Jump to search

Examples of Use of Fundamental Theorem on Equivalence Relations

Arbitrary Equivalence on Set of $6$ Elements: $1$

Let $S = \set {1, 2, 3, 4, 5, 6}$.


Let $\RR \subset S \times S$ be a relation on $S$ defined as:

$\RR = \set {\tuple {1, 1}, \tuple {1, 2}, \tuple {1, 3}, \tuple {2, 1}, \tuple {2, 2}, \tuple {2, 3}, \tuple {3, 1}, \tuple {3, 2}, \tuple {3, 3}, \tuple {4, 4}, \tuple {4, 5}, \tuple {5, 4}, \tuple {5, 5}, \tuple {6, 6} }$


Then $\RR$ is an equivalence relation which partitions $S$ into:

\(\ds \eqclass 1 \RR\) \(=\) \(\ds \set {1, 2, 3}\)
\(\ds \eqclass 4 \RR\) \(=\) \(\ds \set {4, 5}\)
\(\ds \eqclass 6 \RR\) \(=\) \(\ds \set 6\)


Arbitrary Equivalence on Set of $6$ Elements: $2$

Let $S = \set {1, 2, 3, 4, 5, 6}$.


Let $\RR \subset S \times S$ be an equivalence relation on $S$ with the properties:

\(\ds 1\) \(\RR\) \(\ds 3\)
\(\ds 3\) \(\RR\) \(\ds 4\)
\(\ds 2\) \(\RR\) \(\ds 6\)
\(\ds \forall a \in A: \, \) \(\ds \size {\eqclass a \RR}\) \(=\) \(\ds 3\)


Then the equivalence classes of $\RR$ are:

\(\ds \eqclass 1 \RR\) \(=\) \(\ds \set {1, 3, 4}\)
\(\ds \eqclass 2 \RR\) \(=\) \(\ds \set {2, 5, 6}\)