# Fundamental Theorem on Equivalence Relations/Examples/Arbitrary Equivalence on Set of 6 Elements 2

## Example of Use of Fundamental Theorem on Equivalence Relations

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

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

 $\displaystyle 1$ $\mathcal R$ $\displaystyle 3$ $\displaystyle 3$ $\mathcal R$ $\displaystyle 4$ $\displaystyle 2$ $\mathcal R$ $\displaystyle 6$ $\displaystyle \forall a \in A: \ \$ $\displaystyle \size {\eqclass a {\mathcal R} }$ $=$ $\displaystyle 3$

Then the equivalence classes of $\mathcal R$ are:

 $\displaystyle \eqclass 1 {\mathcal R}$ $=$ $\displaystyle \set {1, 3, 4}$ $\displaystyle \eqclass 2 {\mathcal R}$ $=$ $\displaystyle \set {2, 5, 6}$

## Proof

We have that:

$1 \mathrel {\mathcal R} 3$ and $3 \mathrel {\mathcal R} 4$

As $\mathcal R$ is an equivalence relation, it follows that $\mathcal R$ is transitive and so:

$1 \mathrel {\mathcal R} 4$

Thus:

$\eqclass 1 {\mathcal R} \subseteq \set {1, 3, 4}$

but as:

$\size {\eqclass a {\mathcal R} } = 3$

it follows that:

$\eqclass 1 {\mathcal R} = \set {1, 3, 4}$

There are $6$ elements of $S$.

Thus the other $3$ elements must all be in the same equivalence class of $\mathcal R$ which does not contain $1$, for example.

Thus we have:

$\eqclass 2 {\mathcal R} \subseteq \set {2, 5, 6}$

and the information we were given that $2 \mathrel {\mathcal R} 6$ was superfluous.

$\blacksquare$