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.