Intersection of Equivalences
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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: $\S 17 \beta$