# Intersection of Equivalences

Jump to navigation
Jump to search

## Theorem

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

## Proof

Let $\RR_1$ and $\RR_2$ be equivalence relations on $S$.

Let $\RR_3 = \RR_1 \cap \RR_2$.

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 $\RR_3$.

### Symmetric

Equivalence relations are by definition symmetric.

So, by Intersection of Symmetric Relations is Symmetric, so is $\RR_3$.

### Transitive

Equivalence relations are by definition transitive.

So, by Intersection of Transitive Relations is Transitive, so is $\RR_3$.

Thus $\RR_3$ is an equivalence.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Equivalence Relations: $\S 17 \beta$