Intersection of Equivalences

Theorem

The intersection of two equivalence relations is itself an equivalence relation.

Proof

Let $\mathcal R_1$ and $\mathcal R_2$ be equivalences on $S$, and let $\mathcal R_3 = \mathcal R_1 \cap \mathcal R_2$.

Checking in turn each of the criteria for equivalence:

Reflexive

Equivalence relations are by definition reflexive.

So, by Intersection of Reflexive Relations is Reflexive, so is $\mathcal R_3$.

Symmetric

Equivalence relations are by definition symmetric.

So, by Intersection of Symmetric Relations is Symmetric, so is $\mathcal R_3$.

Transitive

Equivalence relations are by definition transitive.

So, by Intersection of Transitive Relations is Transitive, so is $\mathcal R_3$.

Thus $\mathcal R_3$ is an equivalence.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 17 \beta$