Intersection of Equivalences

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Theorem

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


Proof

Let $\mathcal R_1$ and $\mathcal R_2$ be equivalence relations on $S$.

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.

$\blacksquare$


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